IV The four fundamental subspaces (p 90 Strang)
If A is an m (rows) x n (columns) matrix then the four fundamental subspaces are
The column space of A, R(A), a subspace of Rm (vectors with m real components)
The nullspace of A, N(A), a subspace of Rn (vectors with n real components)
The row space of A, i.e. the column space of AT, R(AT) (vectors with n real components)
The left null space of A, N(AT) (vectors with m real components)
The row space of A
For the echelon matrix U
1 3 3 2
U  0 0 3 1
0 0 0 0
the row space has the same dimension as the rank, r, which is 2 for U (2 non-zero rows). If A has
been reduced to the echelon form U, the rows of U constitute a basis for the row space of A.
The nullspace of A
The nullspace of A is defined by solutions to A.x = 0. If we transform A.x = 0 to U.x = 0 by a set
of row operations then it is obvious that solutions to U.x = 0 must be the same as those to A.x = 0
and so the nullspace of A is the same as the nullspace of U. It has dimension n – r (nullity), i.e. the
number of columns minus the rank. A basis for the nullspace can be constructed using the method
illustrated in the previous lecture where variables are sorted into free or basic variables, depending
on whether a particular column contains a pivot or not after reduction to echelon form.
The column space of A
The name range comes from the range of a function, f(x), which takes values,x, within the domain
to the range, f(x). In linear algebra, all x in Rm (the row space + nullspace) form the domain and A.x
forms the range. To find a basis for R(A) we first find a basis for R(U), the range of the upper
triangular matrix obtained from A by row operations. Whenever certain columns of U form a basis
for R(U), the corresponding columns of A form a basis for R(A). Columns of U containing pivots
constitute a basis for U and corresponding columns of A constitute a basis for R(A). In the example
below the first and third columns of U are the basis for the column space of R(U) and the same
columns are a basis for R(A).
1 3 3 2
U  0 0 3 1
0 0 0 0
3 3 2
1

A2
6 9 5
 1  3 3 0
The reason is that A.x = 0 and U.x = 0 express linear (in)dependence among the columns of A and
U, respectively. These sets of linear equations have the same solution, x, and so if a set of columns
of U is linearly independent then so are the corresponding columns of A.
The dimension of R(A) therefore equals the rank of A, r, which equals the dimension of the row
space. The number of independent columns therefore equals the number of independent rows
which is sometimes stated as ‘row rank equals column rank’. This is the first fundamental
theorem of linear algebra.
The left null space of A
If A is an m x n matrix then AT is n x m and the nullspace of AT is a subspace of Rm. Row vectors
yT which are members of the left nullspace have m components and obey,
yT.A = 0.
This equation shows that vectors in the left nullspace are orthogonal to the column vectors of A.
The dimension of the left nullspace plus the dimension of the column space must equal the number
of columns of A. Hence the left nullspace has dimension m – r.
In summary
Column space
Nullspace
Row space
Left nullspace
R(A)
N(A)
R(AT)
N(AT)
subspace of Rm
subspace of Rn
subspace of Rn
subspace of Rm
dimension r
dimension n - r
dimension r
dimension m - r
Orthogonality and orthogonal subspaces
A basis is a set of linearly independent vectors that span a vector space. The basis {wi} is linearly
independent if, for all a which are members of the space V, the expansion of a
a = c1w1 + c2w2 + c3w3 + … + cnwn,
is unique. The basis vectors {wi} may or may not be mutually perpendicular and they may or may
not be normalised. The condition that they be orthonormal is
wTi . wj = ij
In this notation the length of a vector squared is xT.x. In a pair of orthogonal subspaces, every
vector in one subspace is mutually perpendicular to every vector in the other subspace. For
example, if the basis vectors {a, b, c, d} for a 4 dimensional space are mutually orthogonal, then a
plus b and c plus d form bases for two mutually orthogonal subspaces. The fundamental subspaces
of linear algebra are orthogonal and they come in pairs: the first pair is the nullspace and row space
and the second is the column space and the left nullspace. We can see that the row space of A is
orthogonal to the nullspace through the homogeneous equation which defines the nullspace.
A.x = 0
From this we see that every row in A is orthogonal to every vector x in the nullspace. The equation
which defines the left nullspace
yT.A = 0.
shows that every column in A is orthogonal to every vector yT in the left nullspace. Given a
subspace v of Rn, the space of all vectors orthogonal to v is called the orthogonal complement, v┴.
N(AT) = R(A)┴
N(A) = R(AT) ┴
dimensions (m – r) + r
dimensions (n - r) + r
= m number of columns of A
= n number of rows of A
The nullspace is the orthogonal complement of the row space and the left nullspace is the
orthogonal complement of the column space. This is the second fundamental theorem of linear
algebra.
Column space
Row space
R(AT)
xr
R(A)
xr → A.xr = A.x
x → A.x
x = xr + xn
xn → A.xn = 0
0 = null vector
N(AT)
xn
Nullspace
N(A)
Left Nullspace
Action of the matrix A on the column vector x (Fig. 3.4 Strang)
The figure above shows the action of the matrix A on the vector x. In general, the vector x has a
row space component and a nullspace component. The part of x which lies in the row space is
carried into the column space while the part that lies in the nullspace is carried to the null vector.
Nothing is carried to the left nullspace.
VI Inner products and projection onto lines (p144 Strang)
If x and y are two column vectors, the inner (or dot or scalar) product of the two vectors is xT.y.
The condition for two vectors to be orthogonal is that their scalar product be zero. We want to find
a method for creating an orthogonal basis from a set of vectors which is non-orthogonal.
a
b
p
Projection of vector b onto vector a
The figure shows two position vectors a and b and the position vector, p, of the point on the line
from the origin to a that lies closest to b. The line from p to b is perpendicular to a; this is the
vector b – xa, where x is the length of p. It is perpendicular to a so
aT.(b – xa) = 0.
Rearranging this we get
x=
a T .b
a T .a
Since p is parallel to a then p = xa and this can be rewritten as
p = Pa .b =
aa T
.b
a Ta
Note that aaT is the outer product (a dyad) of a with itself and that this is a matrix while the
denominator is a scalar (as it must be). The outer product of a with itself is the projection operator
Pa which projects any vector it operates on onto a. A projection operator P2 is idempotent with P
(P2, P3, …, Pn has the same action as P). P is a square matrix and its nullspace is the plane
perpendicular to a.
Orthogonal bases, orthogonal matrices and Gram-Schmidt Orthogonalisation
In an orthogonal basis every basis vector is mutually perpendicular to every other basis vector. In
an orthonormal basis, in addition to being an orthogonal to other basis vectors, each vector is
normalised. A set of orthonormal vectors will be denoted {qi}. They have the property that
q Ti .q j  δ ij
The matrix constructed from these vectors is denoted Q.



T
Q .Q  














q1T
q T2
q 3T
q T4
q 5T
 q1

  |
  . |

  |
  |
q2
|
|
|
|
q3
|
|
|
|
q4
|
|
|
|
q 5  1
 
|  0
|   0
 
|  0
|  0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0

0
1
QT .Q  1 and QT  Q-1
If a general vector b is expanded in a basis of orthonormal vectors with coefficients, xi,
b  x1q1  x 2q 2  ...  x n q n
Multiply on the left by qT and use orthonormality of the basis vectors to find x1 = q1T.b etc.
Any vector b may therefore be decomposed as
b  q1q1T .b  q 2q T2 .b  ...  qnqnT .b .
and it is obvious that the unit matrix may be written as
q q
i
T
i
1.
i
Here qiqiT are projection operators onto qi (there are no denominators because the qi are
normalised). The coefficients along each q are






b  q1T .b q1  q T2 .b q 2  ...  qnT .b qn
If the q vectors form a basis then this expansion is unique. Orthogonal matrices, whose columns
consist of sets of orthonormal vectors, have the important property that multiplication of a vector by
an orthogonal matrix preserves the length of the vector.
Q.x  x
Q.x  x
2
2
becomes
Q.x T .Q.x  x T .Q TQ.x  x T .x
Inner products and angles between vectors are also preserved
Q.x T .Q.y  x T .Q TQ.y  x T .y
Angles are preserved because the angle between vectors is given by
 x T .y 


x
y


  arccos
The Gram-Schmidt Orthogonalisation process, for a pair of vectors which are non-orthogonal to
begin with, works by subtracting the projection of one vector on another from one of the vectors to
leave a pair of orthogonal vectors. Suppose that these vectors are a and b
- a aT.b
b – a aT.b
b
a
aT.b
Gram-Schmidt orthogonalisation of b to a
We can generate a new vector b – a aT.b from a and b which is orthogonal to a. We can call these
two orthogonal vectors q1 and q2. If a and b are part of a larger set of non-orthogonal vectors then
this process can be continued, trimming off the parts that are non-orthogonal to the qi already
computed. Usually the vectors are normalised at each step so that the {qi} form an orthonormal
basis. The Gram-Schmidt orthogonalisation process can be written as
a
q1 
a


q2 
b'
b'
b'  1  q 1 q 1T .b
q3 
c'
c'
c'  1  q 1 q 1T  q 2 q T2 .c etc


In matrix form the relationship between the non-orthogonal vectors, a, b, c, and the orthonormal
basis {qi} is
a b c q1
| | |   |

 
 | | |   |
q2
|
|
q 3  q1T a q1T b q1T c


| . 0 q T2 b q T2 c
|   0
0 q 3T c
Here the matrix A on the left contains the original non-orthogonal basis in columns, the matrix at
left on the RHS is an orthogonal matrix, Q, and the matrix on the right on the RHS is upper
triangular and denoted R (because U is already taken in LU decomposition). This is the QR
factorisation of A, A = QR.
VII QR decomposition
Householder transformation H
This is a unitary transformation which takes the first column of the original matrix into minus the
first basis vector while retaining the original length (because it is a unitary transformation). A single
Householder transformation produces a vector anti-parallel to one of the basis vectors and so the
transformed matrix has zeroes in all but the first entry of the first column to which the
transformation is applied. If v is the vector v = x1 +  e1 then the Householder transformation is
H 12
vv T
v
2
 1  2uu T u 

v
v
 1  2uu 
 1  2uu  1  2uu 
 1  2uu 1  2uu 
H T H  1  2uu T
T
T
T T
T
T
T
 1  4uu T  4uu T uu T
1
e3
H1x1 = - e1 x1
x1 First column vector of A
e2
e1
The action of a Householder transformation on the first column vector of a matrix
and H is unitary and preserves lengths of vectors and angles between them. The action of H on x1
is
H.x 1  x1  2
vv T
v
2
.x1
 x1  x1   e1 
2x1   e1 
T
x1   e1  2
H.x 1  x1  x1   e1     e1
2x1   e1  .x1
T
x1   e1  2

2x1   e1 
T
.x1
x1   e1  2
.x1  1
2x1T .x1  2 e1T .x1
2 2  2 e1T .x1
2 2  2 e1T .x1


x1T .x1   x1T .e1   e1T .x1   2 e1T .e1  2   x1T .e1   e1T .x1   2
2 2  2 x1T .e1
In order to bring the original matrix to Hessenberg or tridiagonal form, a sequence of unitary
transformations,
U -13 U -12 U 1-1 AU 1 U 2 U 3 ,
1 0
0 H
11
U1  
0 H 21

0 H 31
0
H12
H 22
H 32
0 
1 0 0
0 1 0
H13 
U2  
0 0 H'11
H 23 


H 33 
0 0 H' 21
0 
1 0
0 1
0 
U3  
0 0
H'12 


H' 22 
0 0
0 
0
0 
1
0 

0 H' '11 
0
is applied to the original matrix, A, which brings A to upper Hessenberg form if A is nonsymmetric and to tridiagonal form if A is symmetric.
b.H T 
1 0  a b 1 0   a

.
0 H c d 0 HT  
T 



 H.c H.d.H 
The column H.c contains only one non-zero entry and the row b.HT contains only one non-zero
entry if b = c (i.e. if A is symmetric).
Application of H without the 1 0 row and column does not result in a diagonal matrix.
 
If H1 .A   0
 0
b c
e f 
h i 
 σH11  bH12  cH13 * *
H1 .A.H  
eH12  fH 13
* *

hH 12  iH 13
* *
T
1
We see that multiplication by H1T on the right hand side destroys the zeros produced by H1 in the
first column. Note that this is not the case when the 1 0 row and column are included.
The subroutine DSYTRD factorises a real symmetric matrix A into a tridiagonal matrix, T.
T = Q-1.A.Q
The diagonal elements of T and the first off-diagonal elements are stored in those positions in A on
exit from DSYTRD. The vectors in each Householder matrix Hi (elementary reflectors) vi used to
construct the orthogonal matrix Q contain n-1 (U1), n-2 (U2), etc. non-zero elements and the
remainder are zeros. The first non-zero element of each v is unity and the remaining elements are
stored in the upper or lower triangular part of A not used for T. The subroutine DORGTR can be
used to assemble the orthogonal matrix Q from the elementary reflectors in A.
QR factorisation
In order to employ the QR algorithm to find the eigenvalues of the original matrix A, we need to
bring the matrix T to tridiagonal or upper Hessenberg form. This can be done using Gram-Schmidt
orthogonalisation or using a further Householder transformation. This time a sequence of
Householder transformation is applied on the left of T to generate an upper triangular matrix, R.
Since Hi are only applied on the left, the zeros in a particular column are not undone as they would
be if T were subjected to a sequence of unitary transformations (as was done to bring A to
tridiagonal form).
b'
  1 b c 
  1



H1 .A   0
e f
H 2 .H1 .A   0
2
 0
 0
h i 
0
H 3 .H 2 .H1 .A  Q -1 .A  R
c'
f' 
i' 
  1 b' '
H 3 .H 2 .H1 .A   0
2
 0
0
c' ' 
f' '   R
  3 
Q  H 3 .H 2 .H1   H1-1 .H -21 .H -31  H1T .H T2 .H T3
1
The QR method for obtaining eigenvalues of a matrix
We are now ready to obtain the eigenvalues of A. The eigenvalues of a triangular matrix can be
read off the diagonal – they are simply the diagonal elements of the triangular matrix. This can
easily be demonstrated by expanding the secular determinant for a triangular matrix.
Suppose that a matrix has been brought to QR form
Ao = QoRo
A new matrix A1 is generated by reversing the order of Q and R
A1 = RoQo
Since orthogonal matrices have the property that QQT = 1 and Q-1 = QT
A1 = RoQo = Qo-1(QoRo)Qo = Qo-1AoQo= Q1R1
A2 = R1Q1 = Q1-1(Q1R1)Q1 = Q1-1A1Q1 = Q2R2
Reversing the order of QoRo can therefore be seen to be equivalent to performing an orthogonal
transformation on Ao. The eigenvalues of a matrix are not changed by an orthogonal
transformation. This is a similarity transformation of Ao etc. and this procedure can be iterated until
convergence is reached when An is an upper triangular matrix. The eigenvalues of an upper
triangular matrix can be read off the diagonal. This can easily be demonstrated by expanding the
secular determinant for a triangular matrix.
VIII Eigenvalues and eigenvectors (p243 Strang)
Everything that has been covered so far has had to do with solving the system A.x = b. The other
important equation of linear algebra is A.x =  x where A is a square (n x n) matrix usually
representing an operator, and x and  are one of the n eigenvectors and eigenvalues of A,
respectively.
If none of the eigenvalues is degenerate then each eigenvector is uniquely determined up to a
normalisation constant and mutually orthogonal – if they are normalised then they will also be
orthonormal. If some of the eigenvalues are degenerate then linear combinations of eigenvectors
belonging to the same eigenvalue are also eigenvectors of A (check by substitution). In the case of
degenerate eigenvalues, it is possible to find eigenvectors, corresponding to the degenerate
eigenvalues, which are mutually perpendicular to eigenvectors belonging to the same eigenvalue
and all others.
Eigenvector directions
uniquely determined when all
eigenvalues non-degenerate
Freedom to choose in a plane for two
eigenvectors with degenerate eigenvalues
Freedom to choose eigenvectors within a plane when two eigenvalues are degenerate
If the eigenvector equation is written as
A  λ1.x  0
where 1 is the n x n unit matrix, then we immediately realise that the eigenvector, x, lies in the
nullspace of the matrix A –  1. That is, the eigenvalues, , are chosen so that A – 1 has a
nullspace and the matrix A – 1 becomes singular (if A – 1 has an inverse then the only solution to
the eigenvalue equation is the trivial solution, x = 0). One way of requiring a matrix to be singular
is that its determinant be equal to zero.
A - 1  0
Setting the (secular) determinant to zero and solving the resulting nth order polynomial for values
of  is convenient when the matrix is small (n ~ 3) but for larger n is impractical. Numerical
methods are required then.
Suppose that an n x n matrix A has n linearly independent eigenvectors. Then if these vectors are
chosen to be the columns, si, of a matrix S, it follows that S-1.A.S is a diagonal matrix . To prove
this, form the matrix A.S
s1

A.S  A. |
|

s2
|
|
s 3   λ 1s 1
 
| |
|   |
λ 2s 2
|
|
λ 3 s 3  s1
 
| |
|   |
s2
|
|
s 3  λ 1

| . 0
|   0
0
λ2
0
0
0   S.Λ
λ 3 
which, by inspection is equal to S., because A.s1 = 1s1, etc. and this can be rewritten as S..
Provided that S has an inverse then A.S = S. can be rearranged to
S-1.A.S = 
or
A = SS-1
In the former case, S is the matrix which is said to diagonalise A and in the latter we see a new
decomposition of A in terms of its eigenvectors and eigenvalues. The matrix S is invertible if all n
eigenvectors, which combine to form the columns of S, are linearly independent and the rank of S =
n. Not all square n x n matrices possess n linearly dependent eigenvectors and so not all matrices
are diagonalisable. This prompts the question: Which matrices are diagonalisable? (p290 Strang)
Diagonalisability is concerned with the linear independence of a matrix’s eigenvectors. Invertibility
is concerned with the magnitude of the eigenvalues (zero eigenvalue implies non-invertible).
Diagonalisation can fail only if there are repeated eigenvalues – if there are no repeated eigenvalues
then each of the corresponding eigenvectors is linearly independent and the matrix is
diagonalisable. Suppose two eigenvectors with distinct eigenvalues are not independent:
c1 x1 + c2 x2 = 0
c1 A.x1 + c2 A.x2 = 0
c1 1 x1 + c2 2 x2 = 0
c1 2 x1 + c2 2 x2 = 0
c1 (1 - 2) x1 = 0
c1, c2 not zero
operate with A A.x1 = 1 x1 A.x2 = 2 x2
subtract first equation times 2
(1 - 2) not = 0 so c1 = c2 = 0 and x1, x2 are linearly independent
If one or more eigenvalues is the same (eigenvalues are degenerate) then, provided a set of linearly
independent eigenvectors can be found for the degenerate eigenvalue then the matrix is still
diagonalisable. This is so if, for  p-fold degenerate, the rank of (A – p1) = n – p. i.e. there are p
linearly independent vectors in the null space of A – p1, which can be chosen to be mutually
orthogonal.
In general, a non-symmetric real matrix will have complex eigenvalues and eigenvectors. In
particular, real symmetric and hermitian matrices have real eigenvalues. Eigenvalues are generally
associated with (1) frequencies in classical or quantum mechanical systems, (2) stability of classical
mechanical systems whereby a change in character of the eigenvalue (e.g. crossing zero) indicates
an instability appearing in the system. Since frequencies of real physical systems are generally real
numbers, real systems are represented by real symmetric matrices (classical mechanics) or
Hermitian matrices (quantum mechanics).
Real symmetric and Hermitian matrices (p290 Strang)
A real symmetric matrix is defined by A = AT whereas an hermitian matrix is defined by B = BH.
BH is the Hermitian conjugate of B, defined by
B   B 
*
H
ij
ji
where the * indicates complex conjugation of the matrix element. If x, y are complex vectors ( Cn
the space of complex vectors with n components) then

the inner product of x and y is xH.y and the vectors are orthogonal if xH.y = 0.

the length of x is xH.x

(AB)H = BH AH

if A = AH then xH.A.x is real

every eigenvalue of an hermitian matrix is real

the eigenvalues of a real symmetric or hermitian matrix corresponding to different
eigenvalues are orthogonal to one another.

a real symmetric matrix can be factored into A = Q..QT and an hermitian matrix can be
factored into B = U..UH

U is a unitary matrix which satisfies U.UH = UH.U = 1 c.f. Q.QT = QT.Q = 1 for a real
symmetric matrix. It contains orthonormal eigenvectors of B in columns.

multiplication of a vector by U does not change its length nor does it change the angles
between vectors.
The singular value decomposition (p 443 Strang)
We have seen that a square Hermitian matrix can be factored into B = U..UH. This is a particular
case of a more general decomposition, the singular value decomposition or SVD. Any m x n matrix
can be factored into
C = Q1..Q2T
Where the columns of Q1 (m x m) are eigenvectors of C.CT and the columns of Q2 are eigenvectors
of CT.C (n x n). The r singular values on the diagonal of  (m x n) are the square roots of the nonzero eigenvalues of both C.CT and CT.C.
For positive definite matrices (n x n with all eigenvalues > 0) this is identical to A = Q..QT and for
indefinite matrices (n x n with eigenvalues > 0 or < 0), any negative eigenvalues in  become
positive in  and Q1 differs from Q2. For complex matrices,  remains real but Q1 and Q2 become
unitary. Then C = U1..U2H . Columns of Q1, Q2 give orthonormal bases for each of the four
fundamental subspaces:




First r columns of Q1
Last m-r columns of Q1
First r columns of Q2
Last n-r columns of Q2
column space of C
left nullspace of C
row space of C
nullspace of C
Bases for the four spaces are chosen such that if C multiplies a column of Q2 it produces a multiple
of a column of Q1 C.Q2 = Q1.
To see the connection with C.CT and CT.C, form those matrices
C.CT = Q1..Q2.(Q1..Q2)T
= Q1..Q2.Q2T.T.Q1T
= Q1..T.Q1T
CT.C = Q2.T..Q2T
Q1 must therefore be an eigenvector matrix for C.CT and Q2 must be an eigenvector matrix for
CT.C.
.T is m x n x n x m (m x m) has eigenvalues 12, …, r2 on the first r elements of the diagonal.
T..is n x n and has the same r elements on the diagonal.
The effective rank of an m x n matrix can be established by counting the number of eigenvalues i2
which are less than some small threshold value.
The size of array required to store an image can be greatly reduced by computing the SVD for the
array and its effective rank. By retaining only eigenvectors for which the eigenvalues i2 exceed
the threshold value, the matrix can be reconstructed using a limited set of eigenvectors and
eigenvalues.
Numerical computation of eigenvalues and eigenvectors (p370 Strang)
We will briefly cover the power method and the QR algorithm for diagonalising a matrix. In the
latter approach, the matrix to be diagonalised is brought to QR form by Gram-Schmidt
orthogonalisation. However, this process can be speeded up considerably for large, dense matrices
by first of all applying a transformation which introduces as many off-diagonal zeroes as possible
into the matrix. The Householder transformation reduces the original matrix to upper Hessenberg
form, where non-zero elements appear only in the upper triangular part of the matrix plus the first
diagonal below the diagonal elements of the matrix. In the special case where the original matrix is
symmetric the Hessenberg form is a tridiagonal matrix.
*
*

0

0
B Hess
* * *
* * *
* * *

0 * *
C Tridiagonal
*
*

0

0
* 0 0
* * 0
* * *

0 * *
Once the matrix is in Hessenberg form then that matrix is transformed into QR form by applying a
Gram-Schmidt orthogonalisation.
Extraction of eigenvectors using the inverse power method.
The power method is one which relies on iteration of a vector, u, which contains some part of all of
the eigenvectors, {xi}, (i.e. it is not orthogonal to any eigenvector). Note that a complete set of
eigenvectors form a basis for any problem. Such an eigenvector can be generated as a random
vector by selecting its elements at random.
u   ci xi
i
When uo is operated on by A
A.u o   c i A.x i   c i i x i  u 1
i
i
Repeated iteration results in
A.u k -1   c i ki x i  u k
i

 c1  1
k
n
 n
uk
k


 x 1  c 2  2

 n
k

 x 2  ...  c n x n

k
 
If n is the largest eigenvalue then  i   0 for all except the largest eigenvalue and the random
 n 
vector will converge to the eigenvector with the largest eigenvalue. The factor which determines
how rapidly convergence takes place is |n|/|n-1|. The random vector may have to be scaled in
magnitude at each iteration step. This is the power method for calculating the eigenvector of a
matrix with the largest eigenvalue.
The inverse power method can be used to obtain the smallest eigenvalue of a matrix. This is
similar in spirit to the power method except that it uses the eigenvalue equation in the form
1
x i  A 1 .x i
λi
The iterated series is
A.v k 1  v k or
A -1 .v k -1   d i
i
1
ki
xi  vk

 v k  d 1 x 1  d 2  1
 2
k
1
k


 x 2  ...  d m  1

 m
k

 x m

and if m is the smallest eigenvalue of A then the random vector converges to the eigenvector with
the smallest eigenvalue. The factor which determines how rapidly convergence takes place is
|1|/|2|. Note that the iteration series for the inverse power method requires solution of the system
of linear equations A.vk = vk-1.
In order to obtain eigenvalues between the largest and smallest eigenvalues, we can adopt the
shifted inverse power method. Here we replace A by A –  I so that all eigenvalues are shifted by
the same amount since
A   I.x i  A.x i   I.x i  i   xi
The factor which determines how rapidly convergence takes place is now |1- |/|2 – |. Thus
convergence can be greatly accelerated if  is chosen to be close to one of the eigenvalues. Each
step of the method solves
A - I .w k 1  w k
or
A - I -1 .w k -1   e i
i
1
i   k
xi  wk
If  is close to one of the eigenvalues (precomputed by another algorithm such as QR) then the
algorithm will very rapidly converge to the eigenvector corresponding to the selected value of .
The standard procedure is to factor A into LU and to solve U.x = (1,1,1,…,1) by back substitution.