SEC. 2-1.
47
INTRODUCTION
Assuming a solution of the form
Y2 = A2eic °t
y' = Ale 'wt
(b)
and substituting in (a) lead to the following set of algebraic equations relating
the frequency, co, and the amplitudes, AI, A2:
2
2
(k1 + k 2 )A - k 2A2 = n1co A 1
2co
-k 2 Al + k 2A2 =
Characteristic-Value
Problems and
Quadratic Forms
=
-
)x
+ a,2x -2
0
)X2
0
kL + k2---
(2-1)
where A is a scalar. Using matrix notation, we can write (2-1) as
ax = Ax
(2-2)
or
(2-3)
(a
- 21 2)X
0
-
A1
inYZ1n
+ kjy 2 - k 2 (y2 - Y) = 0
is a hybrid of the
* Also called "eigenvalue" problem in some texts. The term "eigenvalue"
German term Eigenwerte and English "value."
46
712
nrz
A, +
1
-
k2
/'2
=
4A1
(e)
A-2 =A2
The characteristic values and corresponding nontrivial solutions of (e) are
related to the natural frequencies and normal mode amplitudes by (d). Note
that the coefficient matrix in (e) is symmetrical. This fact is quite significant,
as we shall see in the following sections.
Y21
M2
k2
The values of 2 for which nontrivial solutions of (2-1) exist are called the
characteristicvalues of a. Also, the problem of finding the characteristic values
and corresponding nontrivial solutions of(2-) is referred to as a second-order
characteristic-value problem.*
The characteristic-value problem occurs naturally in the free-vibration
analysis of a linear system. We illustrate for the system shown in Fig. 2-1.
The equations of motion for the case of no applied forces (the free-vibration
case) are
m2 y 2 + k2 (y 2 - Y) = 0
m2
(d)
I
and the final equations are
Consider the second-order homogeneous system,
a 2 1x1 + (a2 2 -
c
)2
A = A 12,
A.2 = A 2
INTRODUCTION
(all
A2
We can transform (c) to a form similar to that of(2-1) by defining new amplitude
measures,*
1/1-1I1n
2-1.
2
Fig. 2-1. A system with two degrees of freedom.
Although the application to dynamics is quite important, our primary reason
for considering the characteristic-value problem is that results obtained for the
characteristic value problem provide the basis for the treatment of quadratic
* See Prob. 2-1.
CHARACTERISTIC-VALUE PROBLEMS
48
SEC. 2-2.
CHAP. 2
SECOND-ORDER CHARACTERISTIC-VALUE PROBLEM
Solving (a),
Al = +6
forms which are encountered in the determination of the relative extrema of a
function (Chapter 3), the construction of variational principles (Chapter 7), and
stability criteria (Chapters 7, 18). This discussion is restricted to the case where
a is real. Reference 9 contains a definitive treatment of the underlying theory
and computational procedures.
2
= +1
(2)
-2
al
all -1]
x2 =
°
±
A1
;11,
2-2.
49
= +i
1
-I-
where i =
SECOND-ORDER CHARACTERISTIC-VALUE PROBLEM
By definition, nontrivial solutions of (2-4) exist only when = 21 or 12.
In what follows, we suppose the characteristic values are real. We consider
first the case where a = . Equation (2-4) becomes
We know from Cramer's rule that nontrivial solutions of
(2-4)
(all - ))Xi + a1z2 2 = 0
a2 1 xl + (a2 2 -- )X 2 = 0
are possible only if the determinant of the coefficient matrix vanishes, that is,
I
I
Ia_l l
when
-U12
(2->)
Ia21
a22
(all -
- 0
22
(2-7)
f11 = all + a 2 2
al1 a2 2 - a
2 a2 1
=
al
and the characteristic equation reduces to
2
1A + /32=
The second equation is related to the first by
(all -
= (all - a 2 2)
+ 4(a 12)
-
l)(a2 2 -
1)
- a1 2 a21
X C~~1)
1 011
X(t)
(d)
-~~~~
l
_ all - A
a1 2
X(1 = {x('), x(1 }
Q, = c,{+1 -la a12-
(I)
+
2511
The characteristic equation for this matrix is
+ 6= 0
By definition,
(a)
(e)
and take cl such that (X(1))Tx(l) = 1. This operation is called normalization,
and the resulting column matrix, denoted by Q 1, is referred to as the charac­
teristic vector for A1.
Example 2-1
- (2)(2)= 67
fI = 2
6
(2)(5) - (2)(2)
fl
(c)
where cl is an arbitrary constant. Continuing, we let
2
Since this quantity is never negative, it follows that the characteristic values for a
symmetrical second-order matrix are always real.
= 0
(b)
Since only one equation is independent and there are two unknowns, the solu­
tion is not unique. We define x(?, x21) as the solution for A = Al. Assuming*
that a 2 # 0, the solution of the first equation is
When a is symmetrical, al 2 = a2 1, and
2
a 21
) times the first eq.
Al
This follows from the fact that the coefficient matrix is singular.
second eq. = (
(2-8)
The roots of (2-8) are the characteristic values of a. Denoting the roots by
il, A2,the solution is
(2-9)
42)2
A1 , 2 = (WIt + I-
2
A - 7
(a)
a 2 ,xl + (a2 2 - ; 1l)x 2 = 0
We let
pi - 4
2
-_
Expanding (2-5) results in the following equation (usually called the characteristic equation) for A:
\'-O
2
I-- o}
0
' - (all + a )A + (alla2 2 - a 2 1a 12 )
1f2
l)xl + al 2x
QTQ = 1
*If a2 =O, we work with the second equation.
(2-10)
2
11
(2-11)
50
CHARACTERISTIC-VALUE PROBLEMS
Since Q1 is a solution of (2-4) for
SEC. 2-2.
= Al, we see that
aQ =
Following the same procedure for
=
A2,
SECOND-ORDER CHARACTERISTIC-VALUE PROBLEM
(1)
we obtain
1
a
[2
(2-13)
- 2b2}
The equations for
=1 +
all -+-
-4x, + 2x 2 = 0
22
2x
Also,
Q2Q 2 = 1
2
=
(2-14)
These equations are linearly independent, and the two independent solutions
are
x(l) = {c 1 , 0}
(b)
X(2) = {0, C2 }
The corresponding characteristic vectors are
a 12
-A2)
{1,2}
= 2 = + 1, we find
and
(2)= c2{1, -}
2
1
Q2 = / {1, -} = -/5 {2, - 1}
One can easily verify that
aQj =2
i = 1,2
jQj
QIQ 2 = Q2Q 1 = 0
(2)
(all -- A,)(a
2
X1) = 2X(1 ) = 2c 1
Q, =
Repeating for
= 0
x( )= cl{1, 2}
and the normalized solution is
(2-15)
If a is not symmetrical, there is only one independent nontrivial solution when
the characteristic values are equal.
It is of interest to examine the product, QrQ2. From (2-10) and (2-13),
we have
- x
times the first equation. Solving the first equation,
X(?) = C1
Then,
and
Q = {+1,0}
Q2
{0, + 1}
-c
We see that the second equation is we obtain
i;2Q2
It remains to discuss the case where A1 = iA2. If a is symmetrical, the char­
acteristic values will be equal only when al 1 = a22 and a1 2 = a21 = 0. Equa­
tion (2-4) takes the form
(al - )xl + ()x 2 = 0
(a)
(0)x1 + (all - )x 2 = 0
QTQ2
= +
2
= 1 = + 6 are
at2
aQ
2]
A1 = +6
where
(i\2
a:[1
3]
The characteristic values and corresponding normalized solutions for this matrix are
21 = +5
(a)
2 = -1
15
Now, when a is symmetrical, the-right-hand term vanishes since
{4 , - 1}
Q =
al
- 22 = -(a
22
-
) =
-a
1
all -
(b)
1
and we see that Q Q 2 = 0. This result is also valid when the roots are equal.
In general, QQ2 #¢ 0 when a is unsymmetrical. Two nth order column vectors
U, V having the property that
UT V=VTU = 0
51
Example 2-2
(2-12)
Q1
c2 {1
Q2 =
CHAP. 2
We see that QTQ 2 # 0. Actually,
Q
QTQ 2
=
7
-5
(3)
(2-16)
are said to be orthogonal. Using this terminology, Q1 and Q2 are orthogonal
for the symmetrical case.
= +i
)2 = -i
We have included this example to illustrate the case where the characteristic values are
I IF PROBLEMS
A nrTRI .lTIC.A_\ A
rV,
52
CHAP. 2
SEC. 2-3.
­
I1v-., ,..
cI
(1 - i)xt - 2x 2 = 0
(1 + i)x 2 = 0
Note that the second equation is (1 - i) times the first equation. The general solution is
I-
Repeating for
= C2 {1, W
· ~~-~·L'2
··
When the roots are complex, 22 is the complex conjugate of At,.Now, we take c2 = c.
(
Then, x(2) is the complex conjugate of x 1) We determine c, such that
2
(X( t))Tx( )=
QfQ 2 = Q2Q1 = 0
1
Also, by definition,
Finally, the characteristic values and characteristic vectors are
QtQ
Al,2-- +i
1
= QQ2 = 1
Using these properties, we see that
Q1 2-
q
In general, the characteristic values are complex conjugate quantities when the elements
of a are real. Also, the corresponding characteristic vectors are complex conjugates.
=
SIMILARITY AND ORTHOGONAL TRANSFORMATIONS
and it follows that
f
i
The characteristic vectors for the second-order system satisfy the following
relations:
(a)
aQ1 =AiQ1
aQ
2
=A 2 Q2
Q2] =
LQtJfI [Qt
q-1
2-3.
(2-19)
The matrix operation, p- ( )p where p is arbitrary, is called a similaritytransformation. Equation (2-19) states that the similarity transformation, q-l'( )q,
reduces a to a diagonal matrix whose elements are the characteristic values
of a.
If a is symmetrical, the normalized characteristic vectors are orthogonal,
that is,
= .2, we find
X(2)
=
q-laq
i
{1, 2j}
x(l)
=
(2-20)
qT
A square matrix, say p, having the property that pT
Now, we let
Q2 1
= [Q1 Q21'
1
0
(b)
Example 2-3
(1)
1, ==
[2 2]
Q2]
q = [Q
,
Column j of q contains the normalized solution for )j. We call q the
modal matrix* for a. With this notation, (b) takes the form
aq = qk
(2-17)
2, = +6
Q1 =
- {(1,2}
normalized
A2 =
+1
Q2 =1
{2,-1}
(2-18)
* This terminology has developed from dynamics, where the characteristic vectors define the
normal modes of vibration for a discrete system.
=
p'
is called an
orthogonal matrix and the transformation, pT( )p, is called an orthogonal
transformation. Note that an orthogonal transformation is also a similarity
transformation. Then, the modal matrix for a symmetrical matrix is orthog­
onal and we can write
qTaq =
(2-21)
We can write (a) as
a[Q 1
53
We have shown that the characteristic vectors are always linearly indepen­
dent when a is symmetrical. They are also independent when a is unsymA2. Then, ql # 0 except for the case where a is
metrical, provided that Al
unsymmetrical and the characteristic values are equal. If ql # 0, q-t exists
and we can express (2-18) as
complex. The equations corresponding to ). = ). are
xl-
SIMILARITY AND ORTHOGONAL TRANSFORMATIONS
=[A
q= [Q,
0]i= [+
+0
Q2 = ]= [ 2_ -1]
Q2
CHAP. 2
CHARACTERISTIC-VALUE PROBLEMS
54
We verify that qT = q I and qlaq =
a
2rl
X2
/5
5 2
qTaq
-1]
-1
1]
I [I - 1 126
1 5 2a
_2
01 = [I
[0
5
2
X
= 6
q
0
1]
2-4.
-1|
=
1]
=
IL
Since a is not symmetrical, qT
-
rql//
1
+ a2 X,, = AX 2
+ a2 x 2
4/X/
[l/6
2//7
...
(2-22)
+ an,,,x, = Ax,
ax = Ax
(a -
n
(2-23)
n)x =
In what follows, we suppose a is real.
For (2-23) to have a nontrivial solution, the coefficient matrix must be
singular.
(2-24)
a - I,, = O
-1/17
Q= L//
1/J17
85-
al 2 x 1 + a2 2 x2 +
We can write (2-22) as
q 1. Actually,
0
+ alnXn = AX 1
L 4,- 1
Q2=
[O
allxl + a 2 X2 +
an 1 ,
Q1 =-A={2, + 1}
- 5
-1
J=-I
[
The nth order symmetrical characteristic-value problem involves deter­
minling the characteristic values and corresponding nontrivial solutions for
1
A1, = +5
aq
THE nth-ORDER SYMMETRICAL CHARACTERISTIC-VALUE
PROBLEM
(2)
A
2
55
21
1 r
-1|
0
2
[5
5[0
2]1
21
I r2
v/
THE nth-ORDER SYMMETRICAL CHARACTERISTICS
One can easily verify that
:
2] [1
-1][2
T I[
5[2
(Iq
SEC. 2-4
2 5/3]
- ,/1/31
The expansion of the determinant is
(_)n
(n- _
3n-1
+ p 2,n -2 _
** + (1)
n) = O
where
One can easily verify that
(2-25)
0] = h
I1
q-aq=
q' 10
n
and /3j is the sum of all the jth order minors that can be formed on the diag,An denote the roots, and expressing the characteristic
onal.* Letting Al, A2,
equation in factored form, we see that
(3)
i -21
a
[=L+i
+ ,
fl = A1 + 2 +±
+
... + An-1An
A21A
+
=
A21A
2
3
2
f
fn =
q= 2/3
al
i2 1
[-i
In this case, q involves complex elements. Since the characteristic vectors are complex
conjugates, they are linearly independent and q-' exists. We find q-', using the definition equation for the inverse (Equation (1-50)):
1I2 .i..
q
-Adj
Iqi
2
l
q= 3/4I +
n
We summarize below the theoretical results for the real symmetrical case.
The proofs are too detailed to be included here (see References 1 and 9):
1. The characteristic values A1, A2, ... , Ain, are all real.
. , Qn , are orthogonal:
2. The normalized characteristic vectors Q 1,Q,
QTQj = 6ij
1~lAdjq
I =
(2-26)
(2-26)
i,j = 1,2.
n
j2
* Minors having a diagonal pivot (e.g., delete the kth row and column). They are generally called
principal minors.
CHARACTERISTIC-VALUE PROBLEMS
56
3.
CHAP. 2
The expansion of la - I3} = 0 is
a is diagonalized by the orthogonal transformation involving the normalized modal matrix.
qraq = .
q =
4) = 0
)2
= 3
23 = -1
22 =-3
Writing out ax = 2x, we have
Q1n
QtQ2 *
2)((1-
(3 -
and the roots are
it
where
57
QUADRATIC FORMS
SEC. 2-5.
= [iJj]
(1 - )x
+2x 2
2x
+(1 -
(a)
)x2
Example 2-4
(3 - i,)X 3
=
0
When 2 = 3, (a) reduces to
(1)
a
33
- 922 + 182 - 6 =
+0.42
+ 2.30
23,
+6.28
-
(1 - 2)X3 =
2
I=1,2,3
I - Aj
(2)
O-
j2
L+0.8
+0.68
-0.52
-0.84
+0.54
-0.10
{A2'
=
we
' °}
{0,0,1}
xi
)
- -x
31)
{
Q3
x3' = 0
and
'
2,O]
This example illustrates the case of a symmetrical matrix having two equal characteristic
inde­
values. The characteristic vectors corresponding to the repeated roots are linearly
roots.
repeated
the
for
I
rank
of
is
I13
pendent. This follows from the fact that a -
QUADRATIC FORMS
The homogeneous second-degree function
+ 2a12xlx2 + a2 2 x22
is called a quadratic form in x, x 2. Using matrix notation, we can express
F as
F =a
F
l lx
2 ral
La.2
.0
X3 = C2
2xl + 2x 2 = 0
2x, + 2 2 = 0
4x = 0
for 23 are
vector
characteristic
and
The general solution
2-5.
5
3
)x2
is
Solving the first and third equations for xt and x3 in terms of x2, the general solution
+0.50
=
Q2
=- 1,(a) reduces to
X2
Finally, the modal matrix (to 2-place accuracy) is
-+0.22 +0.51
X2 = C1
= C1
Ql
2x 2
_
=
X
When
_-X = -(3
q = [QtQ2Q3
We see from (b) that (a - 213) is of rank 1 when 2 = 3. The general solution of (b) is
0
To determine the characteristic solutions, we expand ax = 2x,
(5 - )x1
(- 2)x
(b)
0
By specializing the constants, we can obtain two linearly independent solutions for the
repeated root. Finally, the characteristic vectors for L = `2= 3 are
= +11 + 5 + 2= +18
,3 = 5(2) - (-2)(-2) - +6
,,
A,
2
= 0
(O)X3 = 0
2,
and the approximate roots are
2
-2X2 =
2xt
il, fi2,
Since a is symmetrical, its characteristic values are all real. We first determine
25
+9
=
I
5 3+
using (2- ):
The characteristic equation is
f() = 2,3
+2x
- 2x I
01
5 -2
3 - -2
0 -1 I1
a]
a22
{X
x2
xT
58
CHARACTERISTIC-VALUE PROBLEMS
CHAP. 2
SEC. 2-5.
In general, the function
n
F =
n
k=1 j=1
ajkxjxk = x ax
(2-27)
=Fall
F
Lal2
where ajk = akj, for j =A k, is said to be a quadratic form in xl, x2 ,..., x,,.
If F = xTax is nonnegative (2 0) for all x and zero only when x = 0, we call
F a positive definite quadratic form. Also, we say that a is a positive definite
matrix. If F > 0 for all x but is zero for some x :A 0, we say that F is positive
semidefinite. We define negative definite and negative semidefinite quadratic
forms in a similar manner. A quadratic form is negative definite if F < 0 for
all x and F = 0 only when x = 0. The question as to whether a quadratic form
is positive definite is quite important. For example, we will show that an
equilibrium position for a discrete system is stable when a certain quadratic
form is positive definite.
Consider the quadratic form
2
F = blx + b 2x2 +
21 + 22 =
°
'. ' °
xlA
b2
...
X2
0
LO=0
... -"
=
Suppose we specify that
(2-28)
12>0
(b)
Ai >0
A2 >0
(c)
a,
x,
= all
= a
a l la 2
al 2 a1 2
31> 0
b >0
I
Then, letting
y = qTx
x = qy
(2-29)
A =
(2-30)
It follows that F is positive definite with respect to y when all the characteristic
values of a are positive. But y is uniquely related to x and y = 0 only when
x = 0. Therefore, F is also positive definite with respect to x. The problem of
establishing whether xTax is positive definite consists in determining whether
all the characteristic values of a are positive.
(2-31)
2 > 0
(2-32)
f >
f22>
,, > 0
(2-33)
where flj is the sum of all the jth-order principal minors. Equivalent conditions
can be expressed in terms of the discriminants. Let Aj represent the deter­
minant of the array consisting of the first j rows and columns.
(a) reduces to a canonical form in y:
F = xrax = yT[,iij]y
ai
=
The quantities /fj and Aj are called the invariantsand discrimninantsof a.
The above criteria also apply for the nth-oder case. That is, one can show
that a is positive definite when all its invariants are greater than zero.
and at least one of the elements is zero.
Now, to establish whether xrax is positive definite, we first reduce a to a
diagonal matrix by applying the transformation, q-'( )q, where q is the orthog­
onal normalized modal matrix for a. We write
(a)
1
f2 >
A, >0
= (XTq)[.ijijy](qX)
(d)
> 0.
Then, a is positive definite when
b >0
xrax = (xTq)(q- aq)(q- x)
(a)
Jai
al = alla22 - a 2 > 0
Since a > 0, it follows from the second requirement in (d) that a2 2
Therefore, (d) is equivalent to (b). We let
or
b 2 > 0"
=
all >0
It is positive semidefinite when
0
2
#3I>o0
A2 =
bl
a
are equivalent to
When F involves only squares of the variables, it is said to be in canonical form.
According to the definition introduced above, F is positive definite when
b2 > 0,"
al 1 a2 2 -
We see from (a) that the conditions
+ bx,
,
0
2
= all + a2 2
1
A;l2 =
2
bl
al2
a2
Using (2-26), the characteristic values are related by
XJ
bl > 0
59
We consider first the second-order symmetric matrix
E
[XIX ...
=
QUADRATIC FORMS
al1
a1 2
:j12
22
alj
c2j
..
atj
a2
.*
(2-34)
ajj
The conditions,
ik,
N,
I
A1 > 0 A 2 > 0
are sufficient for a to be positive definite.*
* See Ref. 1 for a detailed proof. Also see Prob. 2-15.
A,, >0
(2-35)
CHARACTERISTIC-VALUE PROBLEMS
60
CHAP. 2
6. CRANDALL, S. H.: Engineering Analysis, McGraw-Hill, New York, 1956.
7. NOBLE, B.: Applied Linear Algebra, Prentice-Hall, New York, 1969.
-
Example 2-5
(1)
1
FRAZER, R. A., W. J. DUNCAN and A. R. COLLAR: Elementary Matrices, Cambridge
9.
WILKINSON, J. H.: The Algebraic Eigenvalue Problem, Oxford University Press,
University Press, London, 1963.
London, 1965.
10. FADDEVA, V. N.: Computational Methods of Linear Algebra, Dover Publications,
New York, 1953.
The discriminants are
+1
11.
A2 = 2 -1 = +1
A3 = 1(6-4)- 1(3 -2)
12.
FORSYTHE, G. E., and C. B. MALER: Computer Solution of Linear Algebraic Systems,
13.
Prentice-Hall, New York, 1967.
PETERS, G., and J. H. WILKINSON: "Eigenvalues of AX = BX with Band Symmetric
A and B," Comput. J., 12, 398-404, 1969.
+ 2 + 3 = +6
P2 = (2 - 1) +-(3 - 1) + (6 - 4) = +5
PROBLEMS
f3 = A3 = +1
(2)
1
L
1
Since
A2
1
RALSTON, A., and H. S. WILF: MathematicalMethods for DigitalComputers, Vol. 2,
Wiley, New York, 1967.
+ 1(2-2) = +1
Since all the discriminants are positive, this matrix is positive definite. The corresponding
invariants are
=
8.
1
12 2
1 2 3
A1 =
2-1.
t
b = pTap = p- ap
B = bTb
= pTaTp = p ap
(2-36)
ax = AX
(2-37)
where x = by. Determine the expression for a in terms of A and b.
2-2. Let xl, x2 be two nth-order column matrices or column vectors and
let c1, c2 be arbitrary scalars. If
Clxi + C2 X2 = 0
Now, b and a have the same characteristic values.* This follows from
I,)p = la - I,
(2-38)
Then, if a is positive definite, b is also positive definite. In general, the positive
definite character of a matrix is preserved under an orthogonal transformation.
only when cl = c2 = 0, xl and x2 are said to be linearly independent. It follows
that x1 and x2 are linearly dependent when one is a scalar multiple of the other.
Using (2-10) and (2-13), show that Q1 and Q2 are linearly independent when
;il
= 212-
2-3.
REFERENCES
1. HILDEBRAND, F. B.: Methods of Applied Mathematics, Prentice-Hall, New York,
1952.
2. BODEWIG, E.: Matrix Calculus, Interscience Publishers, New York, 1956.
3. SMIRNOV, V. I.: LinearAlgebra, Addison-Wesley Publishing Co., Reading, Mass.,
1964.
4. TURNBULL, H. W., and A. C. AITKEN: An Introduction to the Theory of Canonical
Matrices, Dover Publications, New York.
5. HADLEY, G.: Linear Algebra, Addison-Wesley Publishing Co., Reading, Mass., 1961.
* See Prob. 2-5.
(b)
where b is nonsingular. Reduce (a) to the form
If a is symmetrical, b is also symmetrical:
AI, = p-'(a -
(a)
By
where A and B are symmetrical nth-order matrices and A is a scalar. Suppose
B can be expressed as (see Prob. 1-25)
2
Suppose b is obtained from a by an orthogonal transformation:
lb-
Consider the system
Ay =
-2
is negative (A2 = -3), this matrix is not positive definite.
b
61
PROBLEMS
(a)
(b)
Determine the characteristic values and the modal matrix for
r3 21
2
7
0
0 5 0
3 0 2
2-4. Following the procedure outlined in Prob. 2-1, determine the charac­
teristic values and modal matrix for
12y, + 12y 2 = 4y,
12y + 63y2 = 9y2
CHARACTERISTIC-VALUE PROBLEMS
62
2-5.
Suppose that b is derived from a by a similarity transformation.
b = p-'ap
Then,
Il = la - )1,l
lb -
and it follows that b and a have the same characteristic equation.
(a) Deduce that
(b) = ,(a)
Let F(i) = 0 be the characteristic equation for a. When the characteristic
values of a are distinct, one can show that (see Ref. 1)
F(a) =
where 0 is an nth-order null matrix. That is, a satisfies its own characteristic
equation. This result is known as the Cayley-Hamilton Theorem.
(a) Verify this theorem for
r2 1
k= 1,2,..., n
3(a)
/3(b) =
Note: F(a) = a2 -
Demonstrate for
The fact that 1I,
x82,
I
1r -2
- 1
I
=
P=
'1
formation is quite useful.
(b)
Show that
a-
(a)
(b)
2-10.
qTaq = -
for n
ia + /3
2 3)
3
F = 2x2 + 4Xlx 2 + 3X2
- 4x1 x 2 + 6xjX 3
F = 3xf + 5X2 + 6
-
8x 2x
3
Show that a necessary but not sufficient condition for a to be positive
:finite is
- 1. Use this result to find the inverse of
dc
a=2
(Hint: Take xi =A0 and xj = 0 for j =#i, j = 1, 2,..., n)
2-11. If a = 0, ax = 0 has a nontrivial solution, say x. What is the
value of x'ax ? Note that 2 = 0 is a characteristic value of a when a is singular.
2-12. Let C be a square matrix. Show that CTC is positive definite when
0 and positive semidefinite when CI = 0.
|CI
(Hint: Start with F = xT(CTC)x and let y = Cx. By definition, F can equal
zero only when x = 0 in order for the form to be positive definite.)
2-13. Consider the product CTaC, where a is positive definite and C is
square. Show that C7aC is positive definite when CI # 0 and positive semidefinite when C I = 0. Generalize this result for the multiple product,
7l
2-7. Positive integral powers of a square matrix, say a, are defined as
2
a
=
aa
2
a 3 = aa
aar =
= aar-1
aa'
If fal # O,a-
(a2 -
=
(c) Establish a general expression for a- ' using (2-25).
2-9. Determine whether the following quadratic forms are positive definite.
2-6. When a is symmetrical, we can write
Express a- in terms of q and
la -+ 212.
(b) Show that
2 31
/3,n are invariant under a similarity trans­
..
63
PROBLEMS
CHAP. 2
exists, and it follows from the definition that
a-lar = ai' -l
(a) Show that a' is symmetrical when a is symmetrical.
(b) Let Ai be a characteristic value of a. Show that ,' is a characteristic
value of ar and Qi is the corresponding characteristic vector.
arQi = XrQi
Hint: Start with aQi = iQi and premultiply by a.
2-8. A linear combination of nonnegative integral powers of a is called a
polynomial function of a and written as P(a). For example, the third order
polynomial has the form
3
P(a) = c0 a
a22
ca
all > 0, a 2 2 > 0,...,ann >
C Cn-
C
laC ,,-1
I
2-14. Let a be an mth-order positive definite matrix and let C be of order
m x . Consider the product,
b = CT aC
Show that b is positive definite only when the rank of C is equal to n. What
can we say about b when r(C) < n?
2-15. Consider the quadratic form
all
F[X1 X2
x]L
1
12
al
a
a1 2
a22
..
' xi
a2n
X2
ann-
X,
+ c2 a + c3 In
+
Note that P(a) is symmetrical when a is symmetrical.
aln
a2n*
,u
knrlxv
64
o A rTI:ITl'.-\ALuIE
?[ IV
I L- IV
65
PROBLEMS
CHAP. 2
PROBLEMS
.......
Hint: Use the result of Prob. 1-23. Verify for
We partition a symmetrically,
(pxp)
r[All
[X 1X2 AT
F
x
(qp)
(pxq)
(p
A1 2
(q
1)
form
where q = n - p. The expansion of F = XTaX has the
F = X A, 1 X
3 0 0
0 2 1
0 0 5 21
matrices,
2-18. Suppose we express a as the product of two quasi-triangular
for example,
2
X2
A22
(q x q)
1)
+ 2XTA 12 A2 + X2A 2 2 X
2
(pxp)
(nxn)
Now, we take X2 = 0 and denote the result by Fp:
Fp = XTA 1X1
G 21
p = 12,...,n
that it
are necessary conditions for a to be positive definite. Note
conditions.
remains to show that they are also sufficient
(b) Discuss the case where Ap = 0.
2-16. Refer to Prob. t-25. Consider a to be symmetrical.
product of nonsingular
(a) Deduce that one can always express a as the
positive definite.
is
a
when
matrices
triangular
lower and upper
(b) Suppose we take
bll = b22 =
bn,,
+1
Show that a is positive definite when
j = 1,2,...,n
gjj > 0
and positive semi-definite when
j =1, 2,. . . , n
gjj > 0
(c)
and at least one of the diagonal elements of g is zero.
Suppose we take g = b. Then,
GIlI B IIJ
and
i\ =
lb 2 .
= bb 1=22
11
(qxp)
Since /A l1 is equal
For Fp > 0 for arbitrary X 1, All must be positive definite.
that A lll must be
to the product of the characteristic values of All, it follows
positive.
(a) By taking p = 1,2,..., n, deduce that
Ap = IAlll > 0
[G
b2PP
when a is
Show that the diagonal elements of b will always be real
positive definite.
partitioned, the
2-17. If a quasi-diagonal matrix, say a, is symmetrically
that
Establish
matrix.
submatrix All is also a quasi-diagonal
i, j = 1, 2, . . , N
a = [Ai
positive definite.
is positive definite only when Ai (i = 1, 2, . ., N) are
where p
(pxq)
0
G 2 2j
1
B1 2
B2 2 J
(qxq)
q = n. We take
B2 2 ' Iq
B 11 = Ip
for arbitrary p when
Show that the diagonal submatrices of g are nonsingular
a is positive definite.