# Appendix

```APPENDIX
Review of matrix theory
A.I Terminology
A system of m linear algebraic equations in n unknowns JCI, x2, . . . , xn, namely,
011*1 + 012*2 H
\~ alnxn
alxxx + a22x2 H
h a2nxn = y2
am2x2
= yi
amnxn =
may be represented by the matrix equation
y\
01/1
021
022
0ml
0m2
•**
a
mn
X
n
[ Jm _
which in turn is represented by the simple matrix notation
Ax = y.
Definition A.I A rectangular array of real or complex numbers is called
a matrix.
In the above example A = [atj] is the matrix of coefficients of the equations.
A matrix having just one column is called a column matrix (or vector), and if it
has one row it is called a row matrix. The entries atj in the array are called the
elements of the matrix. The subscripts identify the location of each element in the
array where the first subscript refers to the row number and the second subscript
refers to the column number.
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Appendix: Review of matrix theory
A matrix with m rows and n columns is called an m by n matrix, written a s m x n .
If m = n the matrix is square and is said to be of order n. For square matrices the
elements which have identical subscripts (i.e., an, an, #33* • • • &gt; ann) constitute the
main or principal diagonal.
A.2 Equality
Definition A.2 Two m x n matrices A and B are equal if and only if
corresponding elements are equal; that is,
A= B
if and only if a(j = btj for all i, j .
Note: This requires that both A and B have the same number of rows and columns.
Definition A.3 Given two mxn matrices A = [atj] and = [btj] we define
A + B = C where C = [ctj] is the m x n matrix whose elements are
Ctj =au +btj
for all i, j .
Example A.I
2 1 2 1 . [1 4 0 1 _ [&quot;3 5 2 &quot;
1 6 5 J &quot; |_3 9 9
Note: Two matrices can be added only when they have the same number of rows and
the same number of columns. Such matrices are called conformable for addition.
A.4 Multiplication by a scalar
Definition A.4 The scalar multiple of an m x n matrix A = [atj] by a
scalar k is the m x n matrix B = [btj] where btj = katj for all i, j .
Example A.2
2
'12 3 1
[4 5 6 J
[2 4
L8 1 0
6
12
This definition is suggested by the rule of addition; that is,
p
2 3 1
4 5 6
p
2 3 1
4 5 6
ri 2 3 1
4 5 6
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p
*
e
8
10
12
A.5 Matrix multiplication
223
A.4.1 Subtraction
This is defined by the rule of addition and scalar multiplication by —1. Thus
C = A - B = A + (-1)5
or
ctj = atj - bu
for all i, j .
A.4.2 Commutative and associative laws for addition and scalar multiplication
The following relations are easily proved from the definitions.
1 A + B = B + A (commutative law for addition).
2 A + (B + C) = (A + B) + C (associative law for addition).
3 (kx + k2)A = k\A + k2A (multiplication by a scalar is distributive with
4 k\(A + B) = k\ A + k\B (scalar multiplication is distributive over matrix
5 ki(k2A) = (kik2)A (associative law for scalar multiplication).
A.5 Matrix multiplication
Definition A.5 IfA = [atj] is anmxn matrix and ifB = [b k] is annxr
matrix, the product C = AB is the m x r matrix C = [cik] where
Cik =
i,k.
z^atjbjk
Example A.3
'-1
2 3] 3
2
4
0
3
=[11
13].
Note: Two matrices can be multiplied only if the number of columns of the first
matrix (A) equals the number of rows of the second matrix (B). In this case A and B
are called conformable for multiplication. If C = AB we say that A premultiplies
B to give C, or B postmultiplies A to give C. The product has the same number of
rows as the first matrix (A) and the same number of columns as the second (B).
The general form of matrix multiplication is illustrated below.
-&quot;
an
ai2
--•
b7i
ain
••• fc*
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Appendix: Review of matrix theory
This definition of matrix multiplication is suggested by the procedure of substituting
variables in a set of linear equations.
Consider
+ 012*2 + 013*3 = h
a2\X\ + CI22X2 + ^23*3 — k2.
(A.I)
If we make a change from x variables to y variables, where
x\ =bnyx
+ bny2
x2 = b2lyx + b22y2
(A.2)
*3 = b3lyX +^32^2
then substituting (A.2) into (A.I) we get
bl2y2) + cLU(b2iyi + b22y2) + aufaiyi
+ b32y2) = kx
bl2y2) + a22(b2iy{ + b22y2) + ci23(b3iyi + b32y2) = k2.
Combining terms in each equation gives
(aubn + anb2l + al3b3l)y{ + (anbu + anb22 + al3b32)y2 = k{
(a2ibn
+ a22b2i + ^23^31)^1 + (021^12 + 022^22 + ci23b32)y2 = k2.
(A.3)
Call the matrix of coefficients of this new set of equations
C = [ctj] where ctj = anbij + ai2b2j + ai3b3j
is the same as defined by the product rule of AB\ therefore C has been formed from
A and B by a process called multiplication.
Observe that the set of equations (A.I) could be written as AX — K by using
the product rule. The equations (A.2) could be written as X = BY which gives
ABY = K, so C = AB is the coefficient matrix of (A.3).
A.5.1 Properties of multiplication
a Three basic properties of scalar multiplication do not hold for matrix multiplication
1 AB ^ BA (matrix multiplication is not in general commutative).
2 AB = &copy; does not imply that either A or B equals 0 , where we define the
zero matrix 0 as the matrix whose elements are all zero. (This could also
be written as 0, but this notation is easily confused with the scalar 0.)
3 AB = AC and A / 0 does not imply that B = C.
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225
A.6 Transpose
Illustration: Consider A =
we see that AB =
Also, AC =
0 0
0 0
0
0
2
1 3
,C =
0 0
0
1
2
0 7
0 0
1
. Here
0
0 , and neither A nor B = 0 .
0 0
, so AB= AC but B ^ C.
0 0
&amp; Matrix multiplication is associative (i.e., (AB)C = A(BC))
To see this assume that the matrices are conformable for multiplication, specifically
A ism xn, B isn x p,C is p xr. Then by the definition of matrix multiplication,
the ij element of the matrix (AB)C is
{(AB)C}tj =
where {AB}ik =
k=\
&pound;=l
Therefore
p
i=\
n
k=\ i=i
and similarly
1=1
k=\
&pound;=l k=\
The order of summation can be interchanged (since scalar addition is commutative),
so the elements are equal for all i, j , and therefore the matrices are equal.
c Matrix multiplication is distributive over addition
That is, A(B + C) = AB + AC and also (A + B)C = AC + BC. This is easily
verified as in the proof of (b) by using the definitions of matrix multiplication and
A.6 Transpose
Definition A.6 If A — [au] is anmxn matrix, the transpose of A is the
n x m matrix B — [bij] = AT where bjt = atj for all i, j . (That is, the
rows of A become the columns of B.)
Example A.4
1 3 5
2 4 6
1 2
3 4
5 6
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Appendix: Review of matrix theory
Properties of transposition
1 (AT)T = A.
2 (A + B)T = AT + BT.
3 (kA)T = kAT for any scalar k.
4 (AB)T = BTAT.
A.7 Special matrices
1 A is symmetric if and only if (iff) A = AT (i.e., atj = a^) for all i and y.
2 A is skew-symmetric iff A = —Ar (i.e., atj = —0//) for all / and j . Note
that this requires that the diagonal elements be always zero.
Example A.5
A=
&quot;1 2 4&quot;
2 1 3 is
is ia symmetric matrix.
4 3 0_
&quot;0
B=
-2
4
2
0
-3
-4
3
0
is a skew symmetric matrix.
3 A is Hermitian if A = (A)T = A*, where the overline indicates the complex
conjugate (i.e., atj = ajt). Note: The diagonal elements are always real,
and a real symmetric matrix is Hermitian.
Example A.6
is a Hermitian matrix.
4 A square matrix A = [atj] is diagonal if atj = 0 for / ^ j .
Example A.7
&quot;3 0 0&quot;
0 2 0 is a diagonal matrix.
0 0 1
5 The identity matrix / is a diagonal matrix with ones along the main diagonal
(i.e., / = [&amp;ij] where 8tj is the Kronecker delta). Sometimes this will be
written as /„ (identity matrix of order n) when it is important to indicate
the order.
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Appendix: Review of matrix theory
Property If A and B are conformable for multiplication and if the two
matrices have been partitioned so that their submatrices are conformable
for multiplication, then
AB=\A
[A,
A 2]
= AXBX +
A2B2.
Proof: Suppose Aism x n and B is n x n. If we partition A so that A\ is
m x p then A2 is m x (n — p). For conformability B\ must be p x n and
B2 is (n — p) x n.
A typical element of A B is {AB}tj = YH=i atkbkj- Now, a typical element
of Ai J5i is {AiSi}/y = Ylk=\ aikbkj (since the first p columns of A are A\
and the first p rows of B are #1).
If we call A2 = [atj] and B2 = [htj] then {A2B2}ij = YJtZx&amp;iibijNote, however, that an = aijP+i and by = bp+ij\ therefore, {A2B2}tj =
YTtZx aitp+ibp+ij. Let k = /? + &pound;, then {A 2 5 2 }/ 7 = E L P + i ^ ^ ^ , and so
{A2B2}U
k=\
k=\
k=p+\
QED.
Example A.10
0 2
1 4
0 1
3 &quot; &quot;4 2
2
1 0
2
0
0
1
1
2
8 0
12 2 | 9
___________
[
C1
C2 1
C3
C4 I
. Then
12
0 2
1 4
0 21
[0 6
2 5 + 0 4
2 1
0 1
[0][0
2] = [0
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l ] + [0
0] = [0
0
2
8
9
1].
A.9 Matrix inversion
229
In general, if A and B are conformable for multiplication and are partitioned as
A2\
A22
&quot;-
A2p
Am\
A o ••• A
Bo]
/&gt;??
' •*
IJ'
2n
B =
BPI
B
p 2
•••
B
p
_
where each partition Atj has qj columns and Bjk has qj rows, then the product
matrix
mp
•••
C = AB =
Cln
C21
can be computed from Ctj — Y%=\ AikBkj. Note that the submatrices of the product
matrix are obtained using the rules of matrix multiplication and addition exactly
as the elements of the product matrix are obtained using scalar multiplication and
A.9 Matrix inversion
This operation arises when we are interested in the linear equation AX = Y and
want to know how X can be written in terms of Y. We cannot write X = Y/A
since matrix division is not defined. However, if we can find a matrix B such that
BA — I then premultiplying both sides by B would give X = BY.
Definition A.7 Given a square matrix of order n (i.e., A is n x n), X is
called the inverse of A if and only if AX = XA = In. Then denote X by
Note: Only a square matrix can have an inverse. If A is not square when B is
conformable for premultiplication it will not necessarily be conformable for postmultiplication. For a nonsquare matrix A which is m x n, if AB = Im then B is
called a right inverse, and if B A = In then B is called a left inverse of A.
Theorem A.I Uniqueness Theorem: If an inverse exists it is unique.
Proof: Assume AB = BA = / and also that AC = CA = I (i.e., assume
that two different inverses exist). Then AB = I and premultiplying by C
gives CAB = C but CA = /, so IB = C\ and B = C.
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QED.
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Appendix: Review of matrix theory
Definition A.8 The adjoint of an nxn matrix A = [atj] is the nxn matrix
B = [btj] = AdjA where btj = \Ajt\ = the cofactor of the element ajt
which is (— l)l+j x the determinant of the matrix resulting from crossing
out the jth row and ith column.
Example A. 11
1
-1
1
A =
2 0
3 1
0 2
\An\
\A2i
\A22
JA 1 3 |
\A23
|A,i|
then
\A3i\
|A 32 |
IA33I
where
| A n | = d e t P M =6 |A21| = (-l)det P Jjl = -4
] =2|A, 2 | = (-l)det f
|A22l=det fj &deg; | =2|A 32 | = (-l)det
|Ai 3 | = det T &quot; 1
3
Q]
= 3
_^
j
= - 3 |A 23 | = ( - l ) d e t fj
= 2
|A 33 |=det
If A = [au] is an n x n matrix and if det A ^ 0 then for B = Adj A/ det A we will
show that AB = I. Multiplying A and B gives
\A2l\
• ••
\Ani
a2n
\Au\
\Al2\
\A22\
• ••
\An2
ann
\Aln\
\A2n\
•
an
det A
1
det A
021
Now consider the typical element along the diagonal
an\An\ +ai2\Ai2\ H
\-ain\Ain\
jA \
det A J a
det A
By the Laplace expansion of det A along the / th row of A we see that the numerator
is det A, so j A^j } = 1. Next consider the typical element off the diagonal for
which / / j :
detAj l 7
_ an\Aji\ +ai2\Aj2\
-\
det A
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\-ain\Ajn\
A.9 Matrix inversion
231
Consider a matrix Z which is equal to A except that its /th row is equal to its jth
row. Since this matrix has two identical rows, we know det Z = 0. Expanding
det Z along row j gives
det Z = det
an
al2
021
&lt;^22 • • •
•••
an
ai2
ajX
aj2
aln
a
2 n
din
H
--•
0yn|A &lt;/n | = 0.
ajn
But since row j is identical to row i then ajk = aik for all k, and the expression
above would give det Z = an\ A j { \ + a i 2 \ A j2\-\
\- ain \Ajn\ = 0. Therefore since
an = 1 and atj = 0, we have shown that AB = I.
In the same way we can show that BA = / ; we therefore have the result that
Theorem A.2 If det A ^ Ofor the n x n matrix A, then A&quot;1 (the inverse
of A) is given by
A
.
det A
This is an existence theorem since if det A ^ 0 we can always form A&quot;1, so
det A / 0 ^&gt; A&quot;1 exists.
Theorem A.3 If A is a square matrix of order n, and if det A = 0, then
no inverse exists.
Proof: Assume a square matrix B exists which makes AB = I. Then
det(A&pound;) = (det A)(det B) = det / = 1. Therefore det A ^ 0 which
contradicts the hypothesis of the theorem. That is to say, when det A = 0
then no B can exist which makes AB = / , so no inverse of A exists. QED.
Definition A.9 If A is a square matrix with det A — 0 the matrix is called
singular, and if A is a square matrix with det A ^ 0 the matrix is called
nonsingular.
Theorem A.4 If A is nonsingular and ifB is any square matrix satisfying
AB = I then B = A&quot;1.
Proof: Since A&quot;1 exists we see A~l(AB) = A&quot;1 (I) shows that B = A&quot;1.
QED.
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Appendix: Review of matrix theory
A.10 Linear spaces
Definition A. 10 A linear space S is a set of elements xt and two operations
which satisfy the following conditions.
1 For any two elements xa and xb in S there exists a unique element xa + xb in
S (called the sum ofxa andxb). This addition operation has the properties:
a
~a + Xb — %b H~ Xa
b Xa + (Xb + Xc) = (xa + Xb) + Xc.
2 There exists a unique zero element 0 such that 0 + x = x and Ox = 0.
3 For every x G S there exists an inverse element —x e S such that x +
(-2)= P.
4 For any numbers a and /3 and any x e S the element ax exists and belongs
to S. Also, a(xa +xb) = OLX^a+axjb', a (fix) = (aft)x', (a + t\$)x — ax + fix.
(Note: The scalars are members of afield; i.e., it contains the sum, difference, product, and quotient of any two numbers except that division by zero
is excluded.)
The set of all n-vectors with the usual definition for vector addition and multiplication by a scalar form a linear space denoted by Un.
Definition A. 11 Let S be a linear space and W a nonempty set ofS. W
is called a subspace iff W is a linear space.
Example A. 12
Let S be the linear space of 3 x 1 matrices (i.e., 3-vectors). Let W be all such
vectors with x^ = 0 . It is easy to verify that the elements of W satisfy conditions
1-4 above.
A.ll Linear independence
Definition A.12 Let x\, x2, ..., xr be a set of elements of a linear space S.
An element x is called a linear combination of the elements X\, x2, ..., xr,
iffx can be expressed as x = 21=1 c(Xi for some choice of scalars ct.
Example A. 13
Consider the set JCI =
, then any vector JC =
combination of JCI and x2. A vector x =
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with
0 is not.
is a linear
A. 11 Linear independence
233
Definition A.13 The set of elements X\,x2,...,xr
are called linearly
dependent if there exists a set of constants C\,c2, . . . , cr such that
tXi
= 0,
1=1
and at least one constant is nonzero. If no such set ofc( constants exist, the
elements X\, x2, . . . , xr are called linearly independent.
Example A. 14
Consider
x2 =
x3 =
To see whether this set is linearly independent, we can form c\X\ +c2x2+c3x3
This gives the equations
= 0.
c\ + 2c2 + 0c3 = 0
2ci + 3c2 + c3 = 0
3ci + 4c2 + 2c3 = 0
and has the solution ci = 2, c2 = — 1, C3 = — 1, so we see that the set is linearly
dependent.
Afote: If a set of elements x\, x2,..., Xk contains a linearly dependent subset, then
the entire set is linearly dependent. Also, a set of elements x\, x2,...,
Xk is linearly
dependent iff one of the elements can be expressed as a linear combination of the
other elements.
Definition A.14 A set of elements x\,x2,
... ,xn of a linear space S is
said to span S if any element in S can be written as a linear
of the set Xi, x2, ..., xn.
combination
Definition A.15 A set B of elements of a linear space S is called a basis
for B if
1 The elements of B are linearly independent.
2 The elements of B span the space S.
The number of elements in the basis of a linear space is called the dimension of the
space.
Theorem A.5 Let X\, x2, . . . , xn be a set of elements of a linear space S.
This set is a basis for S if and only if for every xeS
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there is a set of unique
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Appendix: Review of matrix theory
scalars C\, c2, . . . ,cn such that x = ^ &quot; = 1 cih (i-e-&gt; Wany * can be written
as a linear combination ofx\,x2, . . . ,xn in exactly one way).
The unique scalars c\, c 2 , . . . , cn are called the components of x with respect to the
basis.
Proof:
1 (necessity) Let xux2,...,xnbea.
basis and let x e S. Then, by the
definition of a basis, there exists a set of scalars c1? c2,..., cn such that
x = Y%=\ ci*i' Suppose there is another set of scalars kuk2,... ,kn such
that x = Y%=\ kiXi (i.e., suppose the c,-s are not unique). Then
or
i=\
i=\
Since the xt are a basis, they are linearly independent, so the linear combination above can equal zero only if ct — kt = 0 for all /. Thus ct = ki9 so
the c,-s are unique.
2 (sufficiency) Suppose for every x e S there is a unique set of scalars
ci, c2, • • •, cn such that x = Y%=\ ci*i' Then the set spans S. To show that
the set is linearly independent, consider expressing the zero element as a
linear combination of the set. That is, 0 = X^Li ci*i- ^ u t w e know that
the sum YTt=\ O^i = 0&gt; s o by uniqueness ct = 0 for all /. Thus the set
x\, x2, . . . , xn is linearly independent.
We note that any element of a linear space can be represented by its
components with respect to the basis elements. Since these components
are unique, the representation is unique.
QED.
Example A.15
A vector x =
V
e\ = 0 ;
can be written in terms of the basis vectors
&quot;0&quot;
e2 = 1 ; &laquo; -
&quot;0&quot;
0
As x = 1 • e\ + 2 • e2 + 3 • e^ , thus it is represented by the components {1, 2, 3}
with respect to the basis vectors e\, e2,e^. On the other hand, the same vector can
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A. 11 Linear independence
235
be written in terms of the basis vectors
&quot;1&quot;
0 ;
0
T
u2 =
l
0
;
&laquo;3
=
&quot;l&quot;
l
l
As x = —l-ui — \u2 + 3 -^3, so x is represented by the components!—1, —1, 3}
with respect to the basis vectors u\, u2,113.
Note: In an 72-dimensional vector space, the set e\, e2, . . . , en, where et is the /th
column of the identity matrix /„ (i.e., a unit vector with a 1 in the /th position and
zeros everywhere else) is called the standard basis. When a basis is not specified,
the standard basis is assumed to be the basis.
Theorem A.6 A square matrix A of order n has det A / 0 if and only if
the columns of A are linearly independent (i.e., det A ^ 0 &lt;&pound;=&gt; columns
of A are linearly independent).
Proof:
a Necessity (only if)
For A = [a 1 \ a2 j • • • \ an] assume the columns of A are linearly deVcnan = 0). Let us say c\ ^ 0. We know that
pendent (i.e., c\d\ +c2a2 -\
det [cifli I a2 I • • • I an ] = cx det A and also that the determinant does
not change if a multiple of any column is added to any other. Therefore, add
c2a2 + c^a^ + • • • + cnan to c\ax. From our assumption of linear dependence
of &laquo;i, a2, - • •, an we thus get det [0 j a2 \ • • • j an ] = c\ det A = 0.
Therefore, since c\ 7^ Owe have shown that the columns being linearly
dependent implies det A = 0. Thus if det A / 0, the columns cannot be
dependent (i.e., det A ^ 0 —&gt; columns of A are linearly independent).
b Sufficiency (if) Outline of proof.
1 Assume det A = 0.
2 Argue that if det A = 0 it is possible to obtain a zero column by adding
multiples of columns to some column.
3 The manipulations which lead to a zero column give the dependence
relation.
4 Thus det A = 0 —&gt; columns of A are linearly dependent, so if the
columns are linearly independent — • det A ^ 0 .
QED.
Note: A similar theorem can be shown for rows.
Theorem A.7 LetX\ ,x2, ... ,xnbeaset of linearly independent n-vectors.
Ify is any n-vector then it can be written as a linear combination of the xis
in exactly one way (i.e., the xis, span the space yn and are a basis for yn).
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Appendix: Review of matrix theory
Proof: From the n x n matrix X — \x\, x2, . . . , xn]. Then using the rule
for multiplication of partitioned matrices write the equation
C\
y=
X2
Xn]
Now the xis are linearly independent, so det 1 ^ 0 , and X l exists. Therefore c = X~xy is a solution to the equation above, and it is unique since
X~l is unique.
QED.
Theorem A.8 The set ofn-vectors X\, x2, . . . , xk is linearly dependent for
all k &gt; n.
Proof:
1 If x\, x2,..., xn are linearly dependent, then so are xx, x2,..., xk.
2 If x\, X2, -.., xn are linearly independent, then the vector xk can be written
uniquely as a linear combination of xx, x2, . . . , xn, so the set*i, x 2 , . . . , ^&raquo;
is linearly dependent.
QED.
A.12 Rank of a matrix
Definition A.16 The rank of a matrix A is the maximum number of linearly
independent columns of A.
Example A.16
1 2
0
0
4
8 - 1 - 1
3 6 2
2
Here the rank of A is less than 4 since the columns are 3-vectors and there cannot
be 4 independent columns. We can see thatx 2 = 2x\ andx3 = x*. Thereforex\ and
x2 are dependent; JC3 and JC4 are also dependent. There are only two independent
columns of A, so the rank of A is two.
Theorem A.9 Ifrf = the maximum number of linearly independent rows
and r = the maximum number of linearly independent columns of a matrix
A, then r' = r.
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A. 12 Rank of a matrix
237
Proof: Consider the m x n matrix A. Let ax, a2, . . . ,ar be the independent
columns of A and the first r' rows the independent rows of A:
aXn
012
flu
022
02n
01r
det
ar&gt;n
0(r/+l)2
•••
0(r'+l)r
an2
...
anr
an\
1 Let us assume that r' &lt; r. We can consider the matrix A partitioned as
A=
~'
Bl
01/
where the columns of Bx are bt =
for / = 1, . . . , r. Since the bt are
r'-vectors and r &gt; r' the vectors bx,b2, ... ,br are linearly dependent, so
cxbi + c2*2 H
h crAr = 0
and not all ct are 0.
= Bxc = 0
where c ^ 0.
In matrix notation
[bxb2...br]
cr
2 Now let wf = [an, ai2,...,
A =
ain\ = the /th row of A, so
K
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238
Appendix: Review of matrix theory
Since thefirstr' rows are linearly independent, any other row can be written
as a linear combination of the first r' rows. Thus
u] = dXiu\ + d2iuT2 +•
= dJ[Bx:B2]
drfiuTr, = [dXid2i
for/ = r'
This is the same as
ut,
r'+2
= [B3
[Bx
B4] =
B2]&plusmn;D[BX
B2l
dl
We see therefore that B3 = DBX, and since Bxc = 0we see that B3c = 0.
We now have
Bx
c = 0 where c ^ 0
so the columns of
are linearly dependent. These columns are the first
B3
r columns of A w lich by definition of r are linearly independent. So the
assumption that r &gt; r' leads to a contradiction and says that r' cannot be
less than r. In the same way, if we assume that r' &gt; r and use the same
reasoning as above we also get a contradiction, so r' cannot be greater than
r. We thus conclude that r' — r.
QED.
Theorem A.10 If a matrix A has rank r then
1 Among all the r x r square submatrices of A, at least one has a nonzero
determinant.
2 Every square submatrix of order k &gt; r has a zero determinant.
Proof:
1 This follows directly from the previous proof. As above, arrange the first
r rows and columns of A to be independent ones. Since the matrix Bx
was shown to be r x r, if det Bx = 0 then Blc = 0 for some c / 0 ,
and as before we show that this says the first r columns of A are linearly
dependent. Since by definition these columns are linearly independent we
conclude that det #i cannot be zero.
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A. 13 Simultaneous linear equations
239
2 We know that every set of k &gt; r columns (or rows) is linearly dependent,
since by definition r is the maximum number of independent columns (or
rows). Thus if we form a matrix Ak from any k columns of a matrix A we
have that Akc = 0 where c ^ 0 is a ^-vector. Let A k be a k x k matrix formed
from any k rows of Ak. We see that A kc = 0 with c j=. 0. If det A k ^ 0
then A k exists, and this would imply that c = 0. Since c ^ O this say that
det A k = 0.
QED.
A.13 Simultaneous linear equations
A. 13.1 Motivation and definitions
For the three sets of equations shown, can we obtain a solution?
*1 + *2 = 1
X\ — x2 = 0
,
X\ + X2 = 1
*i + X2 = 1
2JCI + 2x2 = 2
xi + x2 = 2
Set (a) has exactly one solution; set (b) has an infinite number of solutions; set (c)
has no solution. In general we would like to investigate the existence of solutions of
a system of m linear equations in n unknowns whether m &gt; n,m = n, or m &lt; n.
Let us write this system as
H
h aXnxn = bx
021*1 + 022*2 H
V a2nxn = b2
0ml*l + 0m2*2 + * * * + 0mn*n = bm.
Any set of values x\, x2,...,
xn (or vector x) which satisfy these equations is called
a solution.
If a system has one or more solutions it is called consistent, otherwise it is
inconsistent. A is called the coefficient matrix and [A j b] is called the augmented
matrix of the system.
If a consistent system has only one solution (i.e., a unique solution) it is called
determinate', otherwise it is indeterminate.
The basic problems which arise in studying a system of linear equations Ax = b
are:
1 Determine whether the system is consistent or inconsistent.
2 If the system is consistent, determine whether it is determinate.
3 If the system is determinate, find its unique solution.
4 If the system is indeterminate, find the set of all solutions.
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Appendix: Review of matrix theory
A. 13.2 Equivalent systems of equations and elementary operations
Definition A.17 The collection of all solutions of a system of equations is
called a solution set.
Note:
1 Inconsistent systems have the empty set as the solution set.
2 Consistent determinate systems have only a single member in the solution
set.
3 Indeterminate systems have more than one member in the solution set.
Definition A.18 Two systems of equations are equivalent if they have the
same solution set (i.e., if every solution of one system is also a solution of
the other).
Now note that if we are given a system of linear equations, we can perform a
number of operations on them which do not change their solutions. Namely,
1 Interchange any two equations.
2 Multiply any equation by a nonzero constant.
3 Add a multiple of one equation to another.
These are called elementary operations. Therefore if we apply a sequence of elementary operations to a system of linear equations we do not change the solution
set, so we get an equivalent system of equations.
A.I3.3 Gaussian elimination
The Gaussian elimination method (or successive elimination) consists or reducing
a system of linear equations to an equivalent system of equations in triangular
form. For example, in upper triangular form, the last equation contains only the
last unknown; the next to last equation contains only the last two unknowns, and
so forth. Thus the equivalent triangular system can be solved for each unknown in
turn.
Example A.17
For the system of equations
x\
3x\
x\
+
+
+
2^2
2x2
x2
+
+
+
3^3
x3
x3
=
=
=
7
5 the augmented matrix is
3
1 2 3 | 7
3 2 1 | 5
1 1 1 1 3
Step 1 Eliminate x{ from the second and third equations to get
+
2x2
— 4x2
x2
+
3JC3
=
7
— 8x3 = —16 with augmented matrix
- 2x3 =
-4
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1
0
0
2
-4
-1
3
-8
-2
7
-16
-4
A. 13 Simultaneous linear equations
241
Step 2 Eliminate x2 from the third equation to get
JCI
+
2x2
+
3x3
=
=
1 2 3 ! 7
1 2 ! 4
0
with augmented matrix
4
0 0 0 i 0
7
These equations, which are in triangular form, are equal to the original set and can
be easily solved. In this example we find that the last equation becomes identically
zero which means that we did not have three independent equations to start with, so
the system is indeterminate. Thus let us choose x3 = c where c is any constant. The
solution is then x2 = 4 — 2c from the second equation and x\ —1 — 2(4 — 2c) — 3c =
— 1 + c for any choice of c. In vector form, the solution for any c is
x =
4
0
+
&quot; 1&quot;
-2
1
which is called the general solution.
Instead of writing out all the equations, we could have applied the elementary
operations directly to the rows of the augmented matrix. These are called the elementary row operations. In our example we have reduced the augmented matrix
to triangular form. That is, the first nonzero entry of the /th row occurs in column
/ (this is called the pivot for the /th row), and all column entries below the pivot
are zero. The form known as row echelon form has all pivots scaled to 1, and all
other column entries equal to zero. The general solution is obtained from the row
echelon form directly without back substitutions.
In our example starting with the original augmented matrix, we get
1 2 3
3 2 1
1 1 i
7
5
3
-
&gt;
•
1 2
0 -4
0 -1
3
-8
—2
7
-16
-4
1 2 3
0 1 2
0 0 0
7
4
0
Note that in Step 1,-3 times the first row was added to the second row and the
first row was subtracted from the third row. In Step 2 the pivot for row 2 was scaled
to equal 1 by dividing by —4 and the result was added to the third row. To get row
echelon form subtract 2 times the second row from the first row of the augmented
matrix after Step 2. This gives
1 0 -1
0 1 2
0 0 0
—
4
0
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242
Appendix: Review of matrix theory
which corresponds to the equations
Xi-X3 =
-1
x2 + 2x3 = 4.
So Xi and x2 can be expressed in terms of x3 = c to give the general solution.
Example A.18
For the equations
X\
2X!
3X!
+
+
+
'1 2
2 4
3 6
2x2
4x2 6x2 —
0
-1
2
0
-1
2
=
-
x3
2x3
x4
=
2X4
=
7
3 the augmented matrix is
29
5&quot;
3
29
Step 1 Subtract 2 times the first row from the second row and 3 times the first row
from the third row. This gives the new augmented matrix
1 2
0 0
0 0
0
-1
2
0
-1
2
5
-7
14
Step 2 Since the pivot for row 2 is zero we could interchange the second and third
columns (this interchanges the positions of the variables x2 and x 3 ). We
will just use the first nonzero element in row 2 as the new pivot and add 2
times row 2 to row 3 and scale the pivot to 1 by multiplying row 2 by — 1.
This gives
1
2 0 0 15'
0 1 1
0 0 0
This happens to be in row echelon form and corresponds to the equations
x\ + 2x2 = 5
x3 + x4 = 7.
The general solution is obtained by choosing x2 = cx and x4 = c2 where
c\ and c2 are arbitrary constants to give
&quot;5&quot;
0
x= 7
0
+
&quot;-2
1
0
0
0&quot;
0
-1
1
C\
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A. 13 Simultaneous linear equations
243
A. 13.4 Fundamental theorem for linear equations
Theorem A.ll The m equations in n unknowns Ax = b are consistent
if and only if the coefficient matrix A and the augmented matrix [A j b]
have the same rank.
Proof: (if) Let rank A = r and suppose aj,a2,... ,ar (the first r columns)
are the linearly independent columns of A. Since r = rank A = rank
[A | b], this says that ax, a2,..., ar, b are linearly dependent, so we can
express b as a linear combination of the vectors ax, a2,..., ar
cx
c2
-c b = cxax + c2a2 H
h crar = [ax \ a2
;
c2
= A
Cr
0
0
Therefore,
0
is a solution, so the equations Ax = b are consistent.
(only if)
be a solution. Then Yi^Cidi = b. Thus 6 is a linear combicn
nation of a i,a 2 , • • • ,0/i so the sets a i, a2 &gt; • • •,&laquo;&laquo; andai,a 2 , • • • ,a n ,6have
the same number of linearly independent vectors in them. Thus the matrices
A and [A j b] have the same number of linearly independent columns, so
they have the same rank.
QED.
Note: A set of homogeneous equations Ax — 0 are always consistent since x = 0
is a solution. This is called the trivial solution.
We note therefore that to test for consistency we must see whether the rank of
A = rank [A \ b]. The most efficient method for determining the rank of a matrix
is to use Gaussian elimination. This will be discussed in detail in a later section.
A. 13.5 Properties of the general solution
Suppose we have a consistent system of linear equations Ax = b with rank A = r.
Assume that the equations are arranged so that the r xr submatrix in the upper left
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244
Appendix: Review of matrix theory
corner has a nonzero determinant. Partition A as shown below:
A =
an
aXr
din
ar\
arr
arn
amr
Bx
B3
B2
B4
a
m(r+l)
1
where det Bx ^ 0 so Bx exists. Let
x =
br +\
Xr+\
X\
b =
and b =
X =
\
bm
_ xn
_xr_
SO
Bxx + B2x = b
and
B3x + B4x = b.
x
Therefore x = Bl [b — B2x], and for any arbitrary choice of xr+i
1,2, . . . , n - r weget
= ct for / =
(A.4)
x = B~\b - B2c]
so the solution
x—
B-l[b-B2c]
satisfies
B2\x
= b
orequivalently
\ B{
B2
b\
x
c
-1
= 0
(A.5)
for any choice of c. We want to show that it also satisfies [B3 B4]x = b.
Since the system is consistent, we know that
rank
B\
B2
b
B3
B4
b
— r.
The first r rows of this matrix are linearly independent since det Bi ^ 0 (i.e., the
rank of [Z?i B2 b] = r, so it has r independent rows). Therefore, every row of
[B3 B4 b] can be written as a linear combination of the rows of [Z?i B2 b].
In other words, if we let jf denote the rows of the augmented matrix [A b] then
we can write
Jn] = \.du i ••• i dri]
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for / = r + 1, . . . , m.
245
A. 13 Simultaneous linear equations
Writing this out gives
dr(r+l)
y
Y\
y;
which says that
ft] .
ft] =
(A.6)
, and using (A.5) we see
Now, postmultiply (A.6) by the column vector
-1
that
BA
= 0
ft
or equivalently
B4
-1
for any choice of c. We have thus shown that the vector x =
(where x is given
by (A.4)) satisfies the equation Ax — b for any arbitrary choice of c, so it is a
solution.
To verify that any solution has this form, let xp be some arbitrary solution, so
Bi
B2
B?,
BA
Therefore Bxxp
b
=
+ B2xp
B7l[bx =
b
= ft, s o x p = Bll[b
— B2xp].
B2c]
Thus the vector
(A.7)
is a general solution of the consistent linear system Ax = b.
A number of points should be noted.
1 Ifr=n (i.e., the rank of A equals the number of unknowns), then there is no
arbitrary vector c, so the solution is unique and the system is determinate.
The only solution is x = Bx~lb where B\ = A and ft —b.
2 For a homogeneous system (i.e., Ax — 0) we see that
a When r = n, the only solution is x = 0.
b When r &lt;n, nontrivial solutions can exist. They are given by
xh =
-B7lB2c
-BX~XB2
for any arbitrary a.
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Appendix: Review of matrix theory
3 When r &lt; n the system is indeterminate. The general solution (A.7) has
the form
+ xh.
x=
We can show that a general solution can also be written as x = xp + xh, where xp
is any particular solution of the system. This is true because this x satisfies Ax = b
and any other solution xa satisfies A(xa — xp) = 0, so xa — xp is a solution to the
homogeneous equation, namely xh. Thus any solution has the formx — xp + xh.
A. 13.6 Rank using Gaussian elimination
In our earlier discussion of the rank of a matrix, we needed to find the largest square
submatrix with nonzero determinant. The rank can actually be determined more
easily by using Gaussian elimination.
Theorem A.12 If a matrix C with rows c\,c\,' • • ,cTmis row equivalent
to a matrix C with rows c\,c\, • • •, cTm, then the ct can be expressed as
linear combinations of the Cj and vice versa.
Proof: This follows from the definition of row equivalence, since the matrix
C can be formed from C by applying only elementary row operations. These
QED.
row operations give the linear combinations.
Theorem A.13 If a matrix C is row equivalent to C, then the rows of C
and C span the same space.
Proof: Since each ct can be expressed as a linear combination of the Cj, any
vector which can be expressed as a linear combination of the Cj can also be
expressed as a linear combination of the Cj. That is, since c( is expressible
as ti = ]C7=i ^0&quot;c/&raquo; a n y v e c t o r ^at can be written as x = Y17=\ a^i c a n
also be written as
x
=
i=\
7=1
7=1
\i=l
Exactly the same holds true when ct is expressed in terms of ct.
QED.
Therefore to determine the rank of a matrix C (which is the same as the number
of linearly independent rows or columns of the matrix), we can get a row equivalent
matrix C in triangular form. The number of linearly independent rows in this form
will be obvious, and it will be the rank of C.
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A. 14 Eigenvalues and eigenvectors
247
Example A.19
Find the rank of the matrix
C =
1
2
-2 -1
3
1 - 1 0
-1 - 2 -
0
1
2
7
6 - 2 - 1
-5 -1
8
1
- 4
-1
3
2
Add appropriate multiples of the first row to the second through fifth rows, in order
to put zeros in the first column below the pivot. This gives
C =
1 2
6 - 2 - 1
0 3
1 2 - 9
-3
0 - 5 - 1 9 14 4
0 2
8 - 6 - 2
0 0 - 1 1
1
Scale the pivot of row 2 to have a value of 1 and proceed as before to add multiples
of row 2 to the higher numbered rows in order to put zeros in the column below the
pivot. Continue in the same way for row 3. At this point we find that the remaining
rows are all zero. This results in
C =
1
0
0
0
0
2 6
1 4
0 1
0 0
0 -1
-2
_3
-1
0
1
-r
-i
-l
0
I
— •
&quot;i
0
0
0
0
2
1
0
0
0
6 -2
4 _3
1 -1
0 0
0 0
-1
-1
-1
0
0
We observe that the first 3 rows of the final matrix are linearly independent (as are
the first 3 columns); thus rank C = 3.
A.14 Eigenvalues and eigenvectors
A. 14.1 The eigenvector problem
Consider the following problem:
Given a square matrix F of order n, find the set of scalars A. and of nonzero
vectors &pound; which satisfy the equation
F&pound; = k&pound;.
(A.8)
In other words, find the vectors which are transformed by F into a scalar multiple
of itself. We allow the scalars and vector components to be complex numbers.
This is called the eigenvector problem. Any scalar for which this problem has a
solution is called an eigenvalue or characteristic value of the matrix F. The nonzero
vector solution is called the associated eigenvector or characteristic vector.
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Appendix: Review of matrix theory
Note that for (A.8) to have a nonzero solution for &pound;, the homogeneous equation
[XI — F]&pound; = 0 must have nontrivial solutions. We know this happens only if
det[A/ - F] = 0.
(A.9)
This equation is called the characteristic equation of the matrix F. When det[AJ — F]
is expanded out, we obtain anrath-degreepolynomial pX in X called the characteristic polynomial
p(X) = det[AJ - F] = Xn + an^Xn~l + an.2Xn~2 + • • • + axX + &lt;x0.
(A. 10)
Thus the characteristic equation
p(X) = 0
(A.ll)
is anrath-degreepolynomial equation that hasraroots (i.e.,rasolutions for X). The ra
roots are the only values of X that make det[AJ — F] = 0 and thus are the only values
of X for which nonzero vectors &pound; can be found. These roots of the characteristic
equation, denoted as Xx, X2,..., kn, are the eigenvalues of F.
Example A.20
F =\&deg;
[I
M so d e t [ A / - F ] = det \
0J
X
[-1
~l\
=X2-l
= (X + l)(X-l)
= 0.
AJ
Thus X = X\ = — 1 and X — X2 = 1 are the two eigenvalues. For X = — 1, the
eigenvector is found by solving the algebraic equations
which gives
&pound;i = —&amp;.
J
The vector &pound;x = k\ \
. f o r any k\ / 0 is thus an eigenvector associated with
X = — 1. For X = 1, we solve
which gives
Thus &pound; 2 = ^2
&pound;i = %2.
for any k2 / 0 is an eigenvector associated with X = 1.
Note: In this example all eigenvectors corresponding to one of the eigenvalues
span a one-dimensional space (i.e., all the eigenvectors corresponding to X\ = — 1
are colinear with
, and all eigenvectors corresponding to X2 are colinear with
r
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A. 14 Eigenvalues and eigenvectors
249
Definition A.19 The space spanned by the eigenvectors corresponding to
an eigenvalue X of a square matrix F is called the eigenspace associated
with X.
Example A.21
For
so
Thus X = 1 is the only eigenvalue, although it is a repeated root of the characteristic
equation. Using X = 1, the eigenvector is found by solving the algebraic equations
o n r&amp;i ro
0
0
which gives &pound;i = an arbitrary number and &pound;2 = 0. The eigenvector associated with
X = 1 is thus &pound; = k \ \ for any k / 0. Here again the eigenspace associated with
X = 1 is one-dimensional.
Example A.22
For
2 0]
2j
Q
\X-2
so
d e t [ X / - F ] = det|^
Q
Therefore X = 2 is the only eigenvalue. Using X = 2, the eigenvector is found by
solving the algebraic equations
0
&deg;J 1&amp;J = L&deg;
and we see that any two-dimensional vector &pound; will satisfy. Thus we can choose
any two linearly independent vectors as the eigenvectors corresponding to X = 2.
The eigenspace associated with X = 2 is two-dimensional.
Definition A.20 For F a square matrix of order n, with characteristic
equation
det[A./ - F] = (X - X^iX
- X2)n2...
(X - Xs)n&deg; = 0
with Xi, X2, . . . , Xs the distinct eigenvalues andn\ + n2 + • —\-n s — n,we
call nt the algebraic multiplicity ofXt in the characteristic equation.
In Example A.20, the algebraic multiplicity of both —1 and 1 was one. In Examples
A.21 and A.22 the algebraic multiplicity of the single eigenvalue was two.
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Appendix: Review of matrix theory
Definition A.21 The geometric multiplicity ofXt is the dimension of the
eigenspace associated with Xiy for a square matrix F of order n, with the
distinct eigenvalues X\, X2, . . . , Xs. We will denote this as gt.
Note: gt = n — rank[F — A.,-/]. In Example A.20, the geometric multiplicity of both
— 1 and 1 was one. In Examples A.21 the geometric multiplicity of 1 was one. In
A.22 the geometric multiplicity of 2 was two. Note that gt &lt; nt for any /.
Theorem A.14 Let F be ann xn matrix and letX\,X2,...,Xsbe
distinct
eigenvalues ofF. Let &pound;( be any eigenvector corresponding to Xt. The vectors
&pound;i&gt; &pound;2* • • •' €s are Nearly independent.
Proof: Let j be the largest integer for which &pound; l 5 &pound; 2 , . . . , &pound;y- are linearly
independent. If j &lt; s then &pound;l9 &pound; 2 , . . . , &pound;j9 &pound;j+l are linearly dependent, so
we can write &pound;j+l as a linear combination of &pound; j , &pound; 2 , . . . , &pound; / in the form
&pound;; + , = &lt; / , &pound; , + 4 * 2 + • • • + &lt; / ; &pound; ;
(A.12)
where not all the dt = 0 since &pound;j+l ^ 0. Now
A&pound;,. +1 = dxA&pound;x + d2A&pound;2 + -&quot;+
djA&pound;j
so
Here X 7+1 ^ 0 because if A,7+1 = 0 then Xt ^ 0 for / = 1, 2 , . . . , j
since the k\, k2,...,
ks are distinct, and since not all dt are zero (A.13)
says the &pound;i, &pound;2* •••&gt;&pound;/ a r e linearly dependent, which is a contradiction.
Therefore we can divide (A.13) by Xj+i to get
Since the representation of gj+l as a linear combination of the linearly
independent vectors &pound;x, ^ 2 , . . . , ^&gt; is unique, comparing (A. 12) and (A. 14)
we see that d;(A,;/A.7+1) = ^ for all i = 1 , 2 , . . . , 7. Since not all d, are
zero there is at least one, say, dk ^ 0, which shows that kk = A.7-+1. This
is a contradiction since the eigenvalues are distinct. Therefore j cannot be
less than s. Since it can never be greater than s it must equal s.
QED.
Theorem A.15 Let F be ann xn matrix and let X\, A2, . . . , ks be distinct
eigenvalues of F. For each Xt, let &pound; n , &pound; /2 , . . . , &pound;ig. be a set of linearly
independent eigenvectors corresponding to Xt. The entire set of vectors.
are linearly
independent.
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A. 14 Eigenvalues and eigenvectors
251
Proof: Assume they are dependent, then
where not all the atj are zero. Let rjt = Y2%iatj^ij^ s o Vi *s either an
eigenvector corresponding to Xt or it is 0. Now if r){ / 0 for some set of
values of /, call this set S. We see then that (A. 15) says that
which says that the set r/i for i e S is linearly dependent, but this cannot
be true since these are eigenvectors associated with distinct eigenvalues.
So S must be empty and all r]i = 0. This says that
^ aij^j = 0 for all /.
If any of these atj / O w e get a contradiction because the sets &pound; n , &pound; /2 , ...,
&pound;igi were defined to be linearly independent, so the atj = 0 for all / and j .
This says that the original set must be linearly independent.
QED.
A. 14.2 Diagonalization of matrices
Definition A.22 Two n x n matrices A and B are said to be similar if a
nonsingular matrix T exists such that T~l AT = B. The matrix T is called
a similarity transformation.
Definition A.23 A square matrix F of order n is called diagonalizable if
it is similar to a diagonal matrix. That is, if a nonsingular matrix T exists
such that
T~lFT = A
(A.16)
where A is a diagonal matrix with diagonal elements X\, X2, . . . ,Xn.
Theorem A.16 A square matrix F of order n is diagonalizable if and only
if it has n linearly independent eigenvectors.
Proof: (if) If there are n linearly independent eigenvectors &pound;i, &pound;2* • • • &gt; &pound;&laquo;&gt;
letr = to, &amp;...&pound;,!.
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Appendix: Review of matrix theory
Then
FT =
A.i
0
.
0
X2
.
0
0
.
0
0
kn
=TA
l
so T FT — A. Thus the diagonalizing matrix T is the matrix having the
ft linearly independent eigenvectors as its columns. The diagonal matrix A
which is similar to F has the eigenvalues of F as its diagonal elements.
(only if) Suppose F is diagonalizable, then there exists a T such that
T~XFT = A so FT = TA which says that Ftt = Xtti for all / =
1, 2 , . . . , ft where t\,t2,... ,tn denote the columns of T. Thus we see Xt is an
eigenvalue of F and ^ is the corresponding eigenvector. Since t\,t2,... ,tn
are the columns of a nonsingular matrix, they are linearly independent.
QED.
Theorem A.17 A square matrix F of order n is diagonalizable if and only
if g( = m for all i (i.e., the geometric and algebraic multiplicity of each
eigenvalue are the same).
Proof: (if) We know that for all the distinct eigenvalues Xu X2, . . . , Xs of
F, we can obtain g\ + g2 + • • • + gs linearly independent eigenvectors
&pound; &gt;&pound; , • • • , &pound;
,&pound; ,&pound; ,-..,&pound;
, . . . , &pound; &gt;&pound;
&gt;•••&gt;&pound;
Now ft! + ft2 + • • • + ns = ft, so when g, = nt for all / this says that
g\ + g2 + - - - + gs = ft, so there are ft linearly independent eigenvectors,
and thus F is diagonalizable.
(only if) When F is diagonalizable, Theorem A. 16 says that F has a set
of ft linearly independent eigenvectors. Call these
Sll&raquo; S12' • • • &raquo; S l p p S21' S22' ' ' * ' S2P2' • • • &gt; Ssl&gt; Vv2' • • • ' €sps
where &pound; n , &pound;/2, • • •&raquo; &pound;*&gt; a r e eigenvectors corresponding to the /th distinct
eigenvalue and /?i + /?2 + • • • + ps = n. Now pt &lt; gt for each /, since gt
is the dimension of the eigenspace associated with Xt. Thus
1=1
1=1
/=1
since ^ &lt; nt for each /. Thus we see that gt = nt for each /.
QED.
Corollary to Theorem A.17 A square matrix F of order n which has n
distinct eigenvalues is diagonalizable.
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253
A. 75 Generalized eigenvectors
This follows directly from Theorem A. 16: since all n eigenvalues are
distinct, the n corresponding eigenvectors are linearly independent, so F
is diagonalizable.
Example A.23
For
&quot;1 0 0
0 2 1
0 0 1
F =
then
det(A7 - F) = (k - lf(k
- 2)
so k = 1 has algebraic multiplicity n\ = 2, and k = 2 has algebraic multiplicity
n2 = 1. For A = 2 solve
-1
0
0
[F -
0
0
0
0 &quot;
1
-1
&quot;0&quot;
0
0
Hi
so &sect;i = 0, &pound;3 = 0, and &sect;2 is arbitrary. Thus
= k\
is an eigenvector
corresponding to k = 2. For k = 1 solve
&quot;0 0 0&quot;
0 1 1
_0 0 0_
=
_*3_
&quot;0&quot;
0
0
so &sect;2 + &sect;3 = 0 and &pound;j is arbitrary. The general solution has the form &pound; = c\
0
1
-1
are two independent eigenvectors
1
-1
corresponding to A = 1. We see that g2 = 2. F can be diagonahzed using the
transformation
T =
0 1 0&quot;
1 0 1
0 0 -1
SO &pound; 2 —
so T~l =
and &pound; 3 = &pound;3
'0 1 1 &quot;
1 0 0
-1
Lo 0
and
T-iFT
=
&quot;2 0 0&quot;
0 1 0
0 1
L
0
A.15 Generalized eigenvectors
We showed that a square matrix of order n can be diagonalized if and only if gt = nt
for all /. When this is not true, we cannot find n linearly independent eigenvectors.
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254
Appendix: Review of matrix theory
Example A.24
A =
&quot;1 1 2&quot;
0 1 3 so det [XI - A] = (X - l)2(X - 2)
0 0 2_
and we see that X\ = 1 has n\ = 2 and X2 = 2 has n2 = 1. Next we find for
X = X{ = 1 that
&quot;0 1 2&quot;
0 0 3
0 0 1
[A
&quot;0&quot;
0
0
=
_?3_
so &sect; 2 = - 2 &sect; 3 and &pound;3 = 0 so &sect; 2 = 0. We find therefore that &pound; = kx
so
there is only one linearly independent eigenvector corresponding to X\ = 1, so
gx = 1 z/z m. For X = X2 = 2 that
-1
0
0
[A -
so &sect;2 =
1
-1
0
2
3
0_
and &pound;i = ^ 2 + 2^3 =
&quot;5&quot;
form &pound; = k2
=
&sect;3
3.
0
0
0_
A general solution of these equations has the
. We see that there is one degree of freedom (i.e., one arbitrary
constant), so #2 = 1.
Since A is third order and has only two linearly independent eigenvectors, it
cannot be diagonalized.
Definition A.24 A vector &pound;\k) is called a generalized eigenvector of rank
k corresponding to eigenvalue Xt if
0
[A - Xj
but
[A - Xi
= 0.
Note: The eigenvectors we have considered until now are of rank 1. Consider the
set of vectors generated as follows:
[A — Xt
= 0 &lt;— the usual eigenvector
_
c(k-\)
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A. 16 Jordan canonical form
255
Now note that for any k &gt; 1,
[A - kjf-'^f
= [A- kjF-Vtf-&quot; = • • • [A -
V]&pound; 2
0.
Thus multiplying (A.17) by [A — X, I] gives
so we see that each vector &pound;f} generated in this way is a generalized eigenvector of
rank k corresponding to eigenvalue A.,-.
Theorem A.18 Eigenvectors of different rank corresponding to Xt are
linearly independent.
Proof: Suppose
Then premultiply both sides by [ktl —
to give
and since [A.,-/ — A ] ^ &quot; 1 ' ^ ^ ^ 0, this means that ck = 0. Continue in this
way to show successively that ck-\ = 0, ck-i — 0 , . . . , Q = 0, so the
vectors are linearly independent.
QED.
A.16 Jordan canonical form
For square matrices A which cannot be diagonalized, a canonical form more general
than a diagonal matrix can be obtained by using as the similarity transformation
a matrix T, obtained from the linearly independent eigenvectors and generalized
eigenvectors obtained from them. The resulting matrix / = T~lAT is called the
Jordan canonical matrix. It has the block diagonal form
J =
&quot;Ji
0
0
Jo
...
0
0
•••
0
0
(A.18)
Js,
where each block Jk is a square {in x tk) matrix with the form
~kk
0
1
kk
0
1
•••
•••
0
0
(A. 19)
Jk =
1
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Appendix: Review of matrix theory
Note that all elements on the main diagonal equal Xk and all elements on the first
superdiagonal (i.e., where the column index is one greater than the row index) equal
1. The order ik of Jk is equal to the highest rank generalized eigenvector which can
be generated from &pound;(kl). There are as many blocks Jk as there are linearly independent
ordinary eigenvectors (i.e., of rank 1), so by this definition s' &gt; s — the number of
distinct eigenvalues. Note that here the Xk are not all distinct. The index k changes
for each linearly independent eigenvector of rank 1. Each distinct eigenvalue A; for
/ = 1 , . . . , s has gt blocks associated with it, where gt is the geometric multiplicity
of Xt. When each distinct eigenvalue has only one independent eigenvector of rank
1 (i.e., gi = 1 for all A./), then sf — s, l{ — rii the algebraic multiplicity of Xt where
For diagonalizible matrices, since there exist n linearly independent eigenvectors,
each Jordan block Jt is of order 1, so the matrix / above is diagonal. In other words,
a diagonal matrix is a special case of the more general Jordan canonical matrix.
We will now show that when we form the similarity transformation T as
T = [&pound;• ,..., €t' , &pound;2 * • • •' &pound;2 2 ' • • •' &pound;S&quot; • • •' &pound; / ' ] '
(A.20)
1}
then T~* A r is in the Jordan canonical form above. Now since [A—/
this means that A&amp; = XkI^} + ^~l\ so
00
A.1 1 0 ••• 0
0 Ai 1 ••• 0
0 0 •••
:::•.:
0 0 ••• A., 1
0 0 • ••• A,
00
00
A 2 1 0 ••• 0
0 A 2 1 ••• 0
•••
•••
0 0 ••• A* 1
0
00
00
0
• • • • A.2
0 0 •••
00
•
=77
so/ = T~lAT.
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•
•
•
A.,. 1 0 • • • 0
0
A,- 1 ••• 0
• • • •*. A.,
257
A. 16 Jordan canonical form
Example A.25
3
-1
1
1
-1
1
Consider the matrix A =
. From
A.-3
0
det[A/-A] = det
-1
1
-1
X - 2 0
1 X - l
= (X-2)(X2-4X+4) = (X-2)3 = 0
we see that X — 2 is the only eigenvalue with algebraic multiplicity a(2) = 3. Now
using [A - 2/]&pound; = 0 (which is the same as [21 - A]&pound; = 0 we get
1 -1
0 0
1 -1
-1&quot;
0
-1_
=
&quot;0&quot;
0
_0_
There is only one independent equation &pound;i — &pound;2 — &pound;3 = 0. Since one equation in
three unknowns gives two degrees of freedom, we can choose any two linearly
independent solutions. Let us choose &pound;1 = &pound;2 = 1&gt;&sect;3 = 0 as one solution to
1
and choose &pound;j = &pound;3 = 1, &pound;2 = 0 as the other
give the eigenvector &pound;(i)] =
1
solution to give a second independent eigenvector &pound;2
=
. We see that the
geometric multiplicity is g(2) = 2.
(2)
To get a generalized eigenvector, let us try to generate one from [A — 2I]&pound;\ =
1 -1
0 0
1 -1
-1&quot;
0
-1
&quot;1
= 1
0_
J3_
We can see that this has no solution, so no higher rank generalized eigenvector can
be generated from &pound;(11). Using [A - 21]^ = ^ \ we find
1 -1
0 0
1 _J
-1&quot;
0
-1
T
=
J3.
0
1
which gives one independent equation, &pound;1 — &pound;2 — &sect;3 = 1- Any solution of this will
give us a generalized eigenvector of rank two. Let us choose &pound;j = 1, &pound;2 = &pound;3 = 0.
Note that any other solution will be linearly dependent on the eigenvectors and
generalized eigenvector already chosen. The canonical transformation we have
found is
&quot;1 1
1 0 0
0 1 o_
r
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Appendix: Review of matrix theory
and
T~lAT =
0
0
1
1
0
-1
0
1
-1
3
0
1
-1
2
-1
-1
0
1
&quot;1 1
r
1 0 0
0 1 o_
=
&quot;2 0 0&quot;
0 2 1 = J
0 0 2_
which is in Jordan canonical form
Theorem A.19 If A is a Hermitian matrix (recall that a real symmetric
matrix is Hermitian) then
1 All eigenvalues of A are real and
2 The eigenvectors &pound;y and &pound;k associated with distinct eigenvalues are orthogonal (perpendicular), that is, &pound;j&pound;^ = 0.
Proof:
1 Let Xt be an eigenvalue and &pound;t its associated eigenvector. Then A&pound;( = Xi&pound;i9
and using &pound;• to denote the complex conjugate of &pound;( we see that &pound;. A&pound;t =
A,&pound; &pound;;. Taking the complex conjugate of this gives &pound;TA&pound;T = Xi&pound;T&pound;t,
and taking the transpose gives &pound;. &pound;t = A.^. ^-. But since A is Hermitian,
AT = A, so we see that X( = Xt which says that Xt is real.
2 Using A&pound;j = A7^7 and A&pound;k = Xk&pound;k where Xj ^ Xk we see that
and
iJ
Now take the complex conjugate, transpose the second equation, and use
AT = A to get
Now if ^ &pound; 7^ 0 the above equation says that A7 = A*, but since k ,• and
-r
A^ are distinct, this is not true, so we see that &pound;k &pound;7 = 0.
QED.
Definition A.25 P is called an orthogonal matrix if P~l = PT.
Theorem A.20 If A is real and symmetric then an orthogonal matrix P
exists such that PT AP is diagonal with diagonal elements equal to the
eigenvalues of A.
Note in the proof that follows we will assume that the eigenvalues of A are
distinct. This is not needed in a more complicated proof of the theorem.
Proof: We know that using T = [&pound;1? &pound;2&gt; * * *&raquo; &pound; J w iU &sect; i y e T~lAT = A.
If we just define the columns of P to bep,- = &pound;//||&pound;/|| (i.e., thep; are unitlength eigenvectors) then P~lAP = A. But sincepjp t = 1 andpjp,- = 0
for j 7^ /, we see that PTP = I, so P r = P &quot; 1 .
QED.
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A. 17 The Cayley-Hamilton theorem
259
A.17 The Cayley-Hamilton theorem
This theorem states a very useful property of square matrices.
Theorem A.21 Every square matrix A satisfies its own characteristic
equation. That is, for p(X) = det(A/-A) = Xn+an-ikn-~1-\
\-aik+a0,
the matrix p(A) defined by
p(A) = An+ an.xAn-x
+ ..- + alA + a0I
equals the zero matrix 0 .
Proof: Let us denote the adjoint matrix Adj[AJ — A] by V(X) and note that
since each element F^-(X) of F(A.) is obtained from the determinant of an
(n — 1) x (n — 1) matrix, each rtj(X) is a polynomial in X of degree no
greater than Xn~l. Thus we can write for T(X)
T(X) = Kn_xXn-x
+ Kn_2Xn~2
+ ..- + KlX + K0
(A.21)
where Kt are (n x n) constant matrices. From
-L&plusmn; = [U-A\
p(X)
we see that
=I
(A.22)
{Kn_xXn-1 + Kn_2Xn~2 + ... + Kxk + K0)[XI - A]
= (Xn + an-ikn-1 + '&quot; + axX + ao)I.
(A.23)
In order for (A.23) to hold for all X the coefficients of each power of X must
be equal. Thus we get
Kn-X = I
Kn-2 — Kn_\A = an-\I
Ko-
KlA =
-K0A = aol.
(A.24)
Now postmultiply the first equation of (A.24) by Aw, the second by An_i,
the third by An_2, and so on, and add all the resulting equations to get
An + an_x An~l + • • • + ax A + aol = 0 .
(A.25)
This is the desired result.
QED.
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Appendix: Review of matrix theory
When all the eigenvectors of A have a rank of one, there is a shorter proof since
= An&pound;{ + an-x An-l&pound;t + • • • + ax A&pound;t + a^t = p(kt)^
= 0.
Therefore p(A)[^{, &pound;2&gt; • • • &gt; &pound; J — &reg;&raquo; a n d since the matrix T = [&pound;l5 &pound;2, • • •&gt;
&pound; J is nonsingular, it follows that p(A) — 0 .
The fact that the rath degree polynomial function p(A) = 0 means that not all
powers of A are linearly independent. In fact, it says that the nth power of A can be
expressed as a unique linear combination of the powers A1 for / = 0, 1, . . . , n — 1.
Namely,
n-\
An = - ^ f l l - A I &quot;
(A.26)
thus
_J2atAi+l
An+i =
= -a^A&quot;
/=0
-
Y^a^A'
1=0
= &pound;flB_,flI-A' - 5 &gt; ; - . - ^ 4 ^a n + 1 ,,A ; .
;=o
i=o
(A.27)
/=o
This shows that A(n+1) can also be expressed as a unique linear combination of the
powers A1 for / = 0 , . . . , n — 1. Proceeding in the same way, we can thus show (by
induction) that for any TV &gt; n, we can express AN as a unique linear combination
of the powers A1fox i = 0 , . . . , n — 1. Therefore.
n—\
AN = Y, oiNjAl
for any N &gt; n.
(A.28)
/=o
We can also use the theorem to obtain an alternate way to calculate the inverse
of A. Just multiply (A.25) by A&quot;1 and isolate the resulting A&quot;1 term to give
,
n-\
A1
where an = 1. Note that the inverse exists if and only if a0 ^ 0. This is to be
expected since a0 equals the product of the eigenvalues which is nonzero if and
only if all the eigenvalues are nonzero.
Example A.26
f o
11
wefindp(A) = A2 + 3A + 2 = 0, so we get A2 + 3A + 2/ = 0.
For A
[—2 —3J
Therefore
A
-i
_
l
&quot;
2
3
~2
_
3
2
1
1
2
0
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261
A. 18 Leverrier's algorithm
A.18 Leverrier's algorithm
As a by-product of the general proof of the Cayley-Hamilton theorem, we have an
algorithm for getting Adj [XI — A], since the equations (A.24) correspond to the
recursion equation
j-i = KjA + cijl
for j = n - 1, n - 2, . . . , 1 with Kn_{ = I.
(A.29)
Thus we have a procedure for iteratively generating the matrices Kj when the
coefficients aj are known. Note: The equation for j = 0 can be considered as a
check on the roundoff error in computation since K-X should be the zero matrix.
It turns out that we do not even have to evaluate det[A/ — A] to get the coefficients
aj since these can also be calculated iteratively. We can show that
dp(X)
= trr(A.)
dX
(A.30)
where tr denotes the trace, defined as the sum of the diagonal elements of a square
matrix. To do this we consider a matrix F(A.) that has elements which depend on a
scalar parameter k, so that
Mk)
F(X) =
/„„(&gt;.)
or in terms of its column/(A.),
- \fi(k)
i /2(A.)
/„(*)].
Then by the rule for differentiation of determinants,
ddetF(X) _
dk
\f,-i(k)
1=1
(A.31)
where/'(A.) denotes the derivative of/-(A.) with respect to A.. Now when F(A.) =
[XI — A], we see that, since A. appears only in the diagonal elements,/'(A.) = et =
a vector with 1 in its ith position and zeros everywhere else. Therefore when the
j'th term in (A.31) is expanded along the ith column we get
rfdet[A./ - A]
dk
(A.32)
i=\
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Appendix: Review of matrix theory
where Ctj (X) is the cofactor of the ij element of [XI — A]. Note that (A.32) involves
only the cofactors of the diagonal elements of [XI — A], and recalling that
Cn(X)
Cn(X)
ln(k)
Cl2(X)
C22(X)
...
...
c2n(X)
...
Cnl(X)
Cn2(X)
cnn(x)_
we see that the sum of the cofactors of the diagonal elements in (A.32) is just the
trace of V(X). This is the result stated in (A.30).
From (A.21) and (A.30) we get
Kn_2X n~2
2a2X
(A.33)
Again, in order for (A.33) to be valid for all X, the coefficients of each power of X
must be equal. Thus
tr Kn_x = n
tr Kn_2 = (n-
1K_I
tr Ko = ax
which for an — 1 corresponds to the recursion equation
tr (Kj-i) = ,/fly
for 7 = n, n — 1, . . . , 1.
(A.34)
Taking the trace of equations (A.29) gives
MKHl =tv{KjA) + naj.
Subtract (A.34) from this to give
O = tr(KjA) + naj - ]a}
or
cij =
:tr(A^ A)
for j = n — 1, . . . , 0.
(A.35)
This iterative procedure, known as Leverrier's algorithm, can be summarized as
follows:
1 Start with Kn_i = I (we know that an — 1). Then for j = n - 1
2 Calculate^ = -^rjtr (KjA),
3 Calculate K^ = KjA + cijl,
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A. 18 Leverrier's algorithm
263
4 Decrease j by 1 and return to (2) if j &gt; 0. (This also gives the check on
* - ! = 0.)
Example A.27
1
0
we find directly that dtt[XI — A] = p(X) =
-2 -3
2
X + 3X + 2. We also find directly that
For the matrix A =
x
2
- i i _ rx + 3
i
A + 3J ~ _
| - 2 A.
Using the algorithm we find
&quot;1 01
0 1
„ A f 0
K\A =
-2
1
-3
= 3
then
-2
0
3 1
-2 0
0
-2
which are indeed the values found directly.
Example A.28
A =
3 0 3
2 4 1
1 1 2
so n = 3, K2 = I, and a3 = 1. We get
a2 = - t r A = - 9
&gt;
0
-5
3&quot;
1
1 -7_
2.
1
Z^ = A - 9 / =
2
'-15
-3
-2
&quot; 7
-3
-2
3
3
-3
i
01 =
- -
=
=
i
~9
tr
3 -12&quot;
-19
3 = 22
- 3 -10_
-12&quot;
3
12
= - | [ 1 5 + 15 + 15] = - 1 5 .
We check the result by finding
K0A + a0I=
15
0
0
0 0
15 0
0 15
-157 = 0
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= 2
264
Appendix: Review of matrix theory
which is correct. Using this result gives the inverse matrix
1
A.3 - 9X2 + 22X - 15
|~A 2 -6A + 7
3
2
2A.-3
X - 5X + 3
X-2
A. - 3
3X- 12
A+ 3
X2 - 7A + 12
A.19 The minimum polynomial
We have seen that the Cay ley-Hamilton theorem states that p(A) = 0 . Thus there
exists a polynomial function p(X) of degree n, which gives the zero matrix 0 when
evaluated at A. It should be apparent that there are other polynomial functions of
the square matrix A which also equal 0 . In fact, for any polynomial function q(A)
the function p(A)q(A) = 0 since p(A) = 0 . Thus there are many polynomials
which equal 0 when evaluated at A. These are called annihilating polynomials for
A.
Definition A.26 The minimum polynomial pm(X) of a square matrix A is
the polynomial of lowest degree m such that pm(A) = 0.
Theorem A.22 IfXj is an eigenvalue of A, then it is a zero of pm(X) (i.e.,
Proof: Let
i=0
i=0
where m is the degree of pm(X). Therefore, for &pound;y an eigenvector of A,
/=o
/=o
0 this tells us that pm(Xj)
Since
— 0.
QED.
Note: If A has n distinct roots, then pm(X) must have n roots which implies that
pm(X) = p(X). Only when A has repeated roots can m be smaller than n.
Example A.29
p
o o o&quot;
4
For A = 0 2 0 0 we find that p(X) = (X — 2) . Since pm(X) must have a
0 0 2 0
.0 0 0 2_
zero at X = 2, we try pm (X) = (X - 2). We can check that pm (A) = A - 2 / = 0 ,
so pm (X) = (X — 2) is correct.
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A.20 Functions of a matrix
265
Example A.30
&quot; 2 1 0 0
0
0
_0
whether /?m(A)
For A =
2 0 0
we find that p(X) = (A — 2)4 as before. Again, to see
0 2 1
0 0 2
= (A — 2), we check A — 2/ and find
0
0
0
0
A-2/ =
1
0
0
0
0
0
0
0
0
0
1
0
so this is not correct. Next let us try pm(X) = (A — 2) 2 . We find that
so pm(A) = (A. — 2) 2 is the minimum polynomial.
A.20 Functions of a matrix
We have seen that polynomial functions of a square matrix can be defined directly.
In Chapter 4 (Section 4.3.3) we also showed how arbitrary functions which can
be expressed as a power series can be defined on matrix arguments. We derived
two methods (Methods 2 and 3) for diagonalizable matrices. In this section we will
derive the generalization of those results, which were stated in Chapter 4 (without
proof), for matrices whose Jordan canonical form is not diagonal.
First note that since the Jordan matrix / in (A. 18) is block diagonal we can easily
see that for any power /
J[
0
1
J\
0
0
0
0
(A.36)
r=
where the order tk of each block Jkl equals the rank of the highest rank eigenvector
that can be generated from &pound;^\ Therefore in the same way as for diagonalizable
matrices, any function which can be expanded in a power series such as /(A.) =
L-, ~ 0 a, A.' can be expressed in block diagonal form as
/(/,)
0
1
f(J2)
•••
•••
0
0
;=o
0
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f(Js')
(A.37)
266
Appendix: Review of matrix theory
where each block /(/*) = X^o at ^l *s -^ x ^ - If we look at a typical block Jk, we
see that
1
0
o xk
1
Xk
2Xk
1
2Xk
Xk
0
0
1
2Xk
0
0
1
1
Xk
o
o
...
...
...
0
1
o
o
o
(A.38)
2Xk
X\
1
2Xk
Note that in (A.38) thefirsttwo superdiagonals above the main diagonal are nonzero.
(If r = the row index of an element and c = the column index then the yth
superdiagonal consists of all elements for which c = r + j.) Multiply J% by Jk to get
X\
2Xk
1
0
0
0
X\
2Xk
1
0
0
0
kl
2Xk
1
...
...
...
o
o
o
2Xk
1
X\
2Xk
•••
•••
3X\
3Xk
1
0
0
X\
3X\
3Xk
1
0
0
0
0
3Xl
3Xk
1
0
0
0
0
Xk
o xk
h
h
0
0
(A.39)
3X2k 3 X t
3X\
0
0
1
3Xk
This procedure can be continued. We note that for any power /, each term on the
main diagonal is k\, each term on the first superdiagonal is dXlk/dXk, each term on
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A.20 Functions of a matrix
267
the second superdiagonal is (1/2!) (d2k[/dk2k), and each term on the third superdiagonal is (l/3\)(d3Xik/dX3k). In general, we see that all terms on the yth superdiagonal
are (l/jWd'KJdki).
Therefore when we form /(/*) = EZoaiJl&gt; w e find t h a t
all terms on the main diagonal are
&gt;A.i =/(A*)
;=o
all terms on the first superdiagonal are
i=0
'dXk
dkk
dXk
li=0
all terms on the second superdiagonal are
d2f(Xk)
1 d1
1,
3;
2!-
2\dX\
L/=o
&quot;2!
dX\
and, in general all terms on the 7 th superdiagonal are
1 djXk _ 1 dj
^
!=0
/ • dXk
dki
J • dXk |_(._0
Thus we find that
i=0
f(h)
f(xk)
0
jj&quot;(xk)
f(xk)
y
i j y,,k,
0
0
0
1
(4_2)!.
f'ih)
where / (v) (-) indicates the vth derivative. Thus f(Jk) is an lk x lk matrix which
can be written as the finite sum
d'f(Xk)
f(Jk) -
(A.40)
7=0
where fi1^ is an 4 x 4 matrix with 1 all along its 7 th superdiagonal and zeros
everywhere else (note that Ek0 is the identity matrix since the Oth superdiagonal is
the main diagonal). Let us rewrite (A.37) as the finite sum
(A.41)
7=0
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268
Appendix: Review of matrix theory
where Ik is a n ^ x w matrix with an lk x lk identity matrix in the k\h block position
and zeros everywhere else. Inserting (A.40) into (A.41) we get
= E E -^r-1!^1* = E E 7^
dA
k=\ j=0
k
k=\ j=0
dk
k
T
where E'kj = (ik) EkjIk.
When we use the canonical transformation T of (A.20), we know that /
T~l AT, so substituting for / in the summation in (A.37) we get
(A.43)
L/=o
;=o
We define the function /(A) for the matrix A as
i=0
so we see from (A.43) and (A.44) that /(A) = Tf(J)T~l.
/ ( / ) from (A.42) into this we get
aK
k=\ j=Q
k
k=\ j=0
When we substitute the
aA
k
This is not quite the equation used in Chapter 4 for Method 2, when A cannot be
diagonalized, because here s' is not the number of distinct eigenvalues.
To get that equation we must combine all the terms corresponding to the same
eigenvalue. Thus let X.x, k2,..., ks denote the distinct eigenvalues, and observe that
we can define Ekj in (A.40) to be the zero matrix when j &gt; &pound;k (i.e., since an &pound;k x lk
matrix has no superdiagonal greater than tk — 1, so Ekj = 0 for j &gt; lk). Thus for
each k corresponding to the same eigenvalue kt we rewrite (A.40) as
*( J\
v^ djf(ki) *
f(Jk) = 2^
j Etj
;=0
dk
(A.46)
i
where mt = max^ corres A( (4) is the size of the largest lk corresponding to kt (Note:
mi is also the multiplicity of k( in the minimum polynomial of A). Then take all
the terms k corresponding to kt to form an nt x nt block diagonal matrix where n{
is the algebraic multiplicity of k{:
7=0
dk
i
where Etj is the nt x nf block diagonal matrix formed from the Ekj for k corresponding to ki. Then following exactly the same steps which lead from (A.40) to
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A.20
Functions of a matrix
269
(A.45) we get
tii-i
n-X
aA
i = \ j=0
i
i = \ j=0
aK
i
where Etj — (7;) r &pound;; 7 /;; lt is an nt x n matrix with an nt x nt identity matrix in the
/th block position and zeros everywhere else. This is the equation used in Chapter
4 for Method 2, when A cannot be diagonalized.
For Method 3 we saw that any function f(X) which is expressible as a convergent
power series, gives a matrix function /(A) which can be represented in terms of
an n — 1 degree polynomial function g(A)
f(A) =
n x
Yn-xA
- + Yn-iA&quot;'2 + • • • + YlA + YoI&plusmn;g(A).
(A.49)
For this representation to be useful we must be able to find the coefficients yt. When
A cannot be diagonalized, we postmultiply (A.49) by the canonical transformation
T of (A.20) and premultiply by T&quot; 1 to get
(A.50)
Since / is block diagonal we find that
f(Jk) = Yn-xK~X + Yn-iJr2 + ' * * + Yx Jk + Yol = gUk)-
(A.51)
f
When the geometric multiplicity gt = 1 for each distinct eigenvalue ki9 then s — s
and &pound;t = nt equals the algebraic multiplicity of A/, and equations (A.40) and (A.47)
become the same, namely,
7=0
d
^i
This equation can be applied to any other function g(X) to give
E
w
7=0
dk
(A-53)
U
i
so we see that if
djf(h)
fori = l,...,s;j
=0,...,ni
(A.54)
then/C/*) = ^ ( / , ) f o r a l l / , s o / ( / ) = g(J). In particular, when g(k) = ^
the n coefficients yo,..., yn-x can be found using the S/=1n/ = n equations given
by (A.54). This is the equation used in Chapter 4 for Method 3, when A cannot be
diagonalized.
When the geometric multiplicity g( is not equal to 1 for some A.,-, then we need
to find fewer coefficients, namely, m the degree of the minimum polynomial. Then
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270
Appendix: Review of matrix theory
m can be used instead of n in (A.49) and mt instead of nt in (A.54). It is usually
not worth the extra effort to find m and the mi9 so it is easier to use (A.54) in this
case also.
A polynomial such as
fix) = 2x\
-lx\
(A.55)
with real coefficients and each term of degree 2 in X[ is called a real quadratic form
in x\ and JC2- In general for n variables x\, . . . , xn, a real quadratic form can be
written as
f(x) = xTAx =
(A.56)
where A is a real n xn matrix. Here we will be concerned only with real quadratic
forms, so we will usually omit the explicit modifier &quot;real.&quot;
Note: When a quadratic form is expressed as in (A.56), many matrices A could
be used to represent the quadratic form. For example, the f(x) = xT Ax in (A.55)
could be represented by any of the matrices
2
2
4
-7
A2 =
2
3
3
-7
We will always take A to be symmetric. That is, we always make atj = a^ =
the coefficient of the xtXj term in the quadratic form, where atj is the entry in the
/th row and jth column of A.
Consider a change of variables from x to y where x — Py and P is nonsingular,
so y = P~lx. Then xTAx = yTPTAPy = yTAy where A is the matrix of the
quadratic form for the new variables y.
Note: When P is chosen as the orthogonal matrix which makes PTAP = A a
diagonal matrix, then the quadratic form becomes yTAy = &pound;&quot;=1A.l-;y? where the
Xt are the eigenvalues of A (this can always be done for a real symmetric A; see
Theorem A.20).
Definition A.27 A quadratic form is called positive definite if it is positive
except when all xt are zero. Thus
x1Ax is positive definite &lt;-&gt; xT Ax &gt; 0
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forx ^ 0.
271
It is called positive semidefinite or nonnegative definite ifxT Ax &gt; Oforx ^
0 (i.e., it is never negative and can equal Ofor a nonzero x).
x\ + x\ is positive definite.
/
1 \2
1
x\ — X\X2 -\—x\ — ( X\
x2 ) is positive semidefinite.
Definition A.28 A real symmetric matrix A is called positive definite if the
quadraticform xT Ax is positive definite (similarly A is positive semidefinite
if its quadratic form is positive semidefinite).
Theorem A.23 A is a positive definite matrix if and only if all the eigenvalues of A are positive. (It is positive semidefinite if all eigenvalues are
nonnegative.)
Proof:
n
x1Ax =yTPTAPy
= yT Ay =
for some orthogonal matrix P. Since x = Py and P is nonsingular, x ^ O
if and only if y ^ 0. For A positive definite the left-hand side is greater than
zero whenever x ^ 0, and therefore the right-hand side is also greater than
zero whenevery -^ 0, which says that all Xt &gt; 0. The same reasoning in the
reverse direction shows that all k( &gt; 0 implies that A is positive definite.
(For semidefinite matrices, the proof is the same except that &quot;greater than
QED.
or equal to&quot; (&gt;) is used instead of &quot;greater than&quot; (&gt;).)
Theorem A.24 If A is positive definite then det A &gt; 0 (if semidefinite then
det A &gt; 0).
Proof: For A positive definite, an orthogonal P exists such that PTAP =
A where A is diagonal with all Xt &gt; 0. Now det A = X\X2 - • • K &gt; 0 and
det P = &plusmn;1 (since (det P)2 = 1). Therefore
det(P r AP) = (det P)2 det A = det A
(For semidefinite replace all &gt; with &gt;.)
so
det A = det A &gt; 0.
QED.
Definition A.29 The kth-order principal submatrix of A, denoted by Ak,
is the k x k matrix formed by deleting the last n — k columns and the last
n — k rows of A.
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Appendix: Review of matrix theory
Example A.31
For the matrix
#n
a 12
#13
#14
#21
#22
#23
#24
#31
#32
#33
#34
_#41
#42
#43
#44 _
the principal submatrices are
Ai=an
A2 =
#11
#12
#21
#22
#11
#12
#13
#21
#22
#23
#31
#32
#33,
and of course A4 = A. Note that the diagonal elements of a principal submatrix
are diagonal elements of A, and the principal submatrices of any Ak are all the
lower order principal submatrices.
Theorem A.25 A real symmetric matrix A is positive definite if and only
if every principal submatrix has a positive determinant.
Note: The determinant of a principal submatrix is called a principal minor, so the
theorem says that A is positive definite if and only if every principal minor is
positive.
Proof: Necessity (only if): when A is positive definite, then
xTAx = ^2^2ciijXiXj
i=l
&gt; 0
7=1
for any x ^ 0. For any choice of k satisfying 0 &lt; k &lt; n, if we make
= xk+2 = • • • = xn = 0 then
k
&laquo;^
JXm\^
k
7
7
LA-} j A-j A&lt; i
^
\J
7=1
i=l
for x\, JC2, . . . , Xk not all zero. Therefore the quadratic form
an
x
k]
•
&gt;0
so the matrix of this form Ak, which is a principal submatrix of A, is positive
definite and det Ak &gt; 0.
Sufficiency (if): we will prove this part by induction. We see that for
n — \ the quadratic form a\\x\ &gt; 0 for JCI 7^ 0 if an &gt; 0; therefore A\
is positive definite. We could go on to show that the theorem is also true
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273
for n — 2. Instead we will assume that we have shown it to be true for
(n — 1) x (n — 1) matrices, and prove that it is then true for the nxn matrix
A.
Form the matrix B — [a\ \ b2 \ b3 j • • • \ bn] from the columns
ak of A by using
for k = 2, . . . , n
(A.57)
Note:
1 We can divide by au since it is greater than zero.
2 B is formed from A by subtracting a multiple of the first column of A from
each of the other columns of A. Therefore each principal submatrix Bk of
B has been formed by subtracting a multiple of the first column of Ak from
each of the other columns of Ak. Since the determinant of a matrix does
not change when such column operations are performed, we conclude that
all the principal minors of A and B are the same.
3 The (n-l)
x (n- 1) matrix
b2n
(A.58)
C =
bnn
is symmetric since (A.57) says that the elements bik are formed by
bik = a[k
an for / = 1, . . . , n, k = 2, . . . , n.
(A.59)
an
Since A is symmetric (i.e., aik = aki) we see that bik = bki for i,k =
2,..., n, so C is symmetric. Equation (A.59) also shows that b\k = 0 for
k =
2,...,n.
4 The operations (A.57) are equivalent to postmultiplying A by the matrix
0
a\\
1
0
0
(A.60)
Q =
0
0
0
0
1
0
0
1
s o B = AQ.
The resulting matrix looks like
~an
a2i
0
b
2 2
•••
•••
0&quot;
b
l n
B =
bn
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Appendix: Review of matrix theory
Next we form another matrix B from B by subtracting multiples of the first
row of B (b\) from the other rows of B (bf) in order to put zeros into the
first column below an.
This is done using
= bj
b\
for / = 2, . . . , n.
(A.61)
an
where bj is the /th row of the matrix B. This is equivalent to B = QT B =
QTAQ for Q given by (A.60) since au = an. We see that since Q is
nonsingular, A is positive definite if and only if B is positive definite. We
will show that B is positive definite.
The resulting matrix looks like.
0
au
0
0
b22
b2n
0
bn2
Kn
B =
Here also, B and B have the same principal minors, so A and B have the
same principal minors which are all positive.
We see that we can express the principal minors of B as
for k = 2 , . . . , n.
det Bk = an
(A.62)
where Ct are the principal submatrices of C in (A.58). Equation (A.62)
shows that all the principal minors of C are positive, because all the principal minors of B are positive and an &gt; 0. Therefore by the induction
assumption, the (n — 1) x (n — 1) matrix C is positive definite. We can
now see that for any n -vector x =
xTBx = anx\+xTCx
l
&gt;0
for any x ^ 0 because au &gt; 0 and C is positive definite. Thus B is positive
definite and so is A.
QED.
Note: Theorem A.25 does not hold for positive semidefinite matrices when all
principal minors are nonnegative. The following matrix is not positive semidefinite
even though the principal minors are nonnegative:
A =
2
0
0
0
0
0
0
0 - 1
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275
Definition A.30 A quadratic form f (x) is called negative definite if—f (x)
is positive definite (f(x) is negative semidefinite if —f(x) is positive
semidefinite).
Note: A is negative definite if and only if (—A) is positive definite.
Theorem A.26 A real symmetric matrix A is positive definite if and only
if a nonsingular real matrix T exists such that A = TTT.
Proof: (if) When T is a nonsingular matrix, then for y = Tx we see that
xTTTTx =yTy &gt; 0 forj ^ 0
but since T is nonsingulary / 0 -^ jr / 0. Therefore xT(TTT)x &gt; 0 for
x ^ 0, so TTT is positive definite.
(only if) When A is positive definite then an orthogonal P exists such that
PT AP = A is diagonal where all Xt &gt; 0. Let
1
0
1
o
0
D =
so DTAD = /. Thus DTPTAPD = /, so A = [DTPT]-l[PD]~l =
([PD]~l)T[PD]-\ and we see that A = TTT where T = [PD]~l is nonQED.
singular.
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```