A First Course in Linear Algebra, 2017a

7.4. Orthogonality 419
insignificant. Therefore, for large m, ⃗u m is essentially a multiple of the eigenvector⃗x n , the one which goes
with λ n . The only problem is that there is no control of the size of the vectors ⃗u m . You can fix this by
scaling. Let S 2 denote the entry of A⃗u 1 which is largest in absolute value. We call this a scaling factor.
Then ⃗u 2 will not be just A⃗u 1 but A⃗u 1 /S 2 .NextletS 3 denote the entry of A⃗u 2 which has largest absolute
value and define ⃗u 3 ≡ A⃗u 2 /S 3 . Continue this way. The scaling just described does not destroy the relative
insignificance of the term involving a sum in 7.7. Indeed it amounts to nothing more than changing the
units of length. Also note that from this scaling procedure, the absolute value of the largest element of ⃗u k
is always equal to 1. Therefore, for large m,
⃗u m =
λ m n c n ⃗x n
S 2 S 3 ···S m
+(relatively insignificant term).
Therefore, the entry of A⃗u m which has the largest absolute value is essentially equal to the entry having
largest absolute value of
( λ
m
)
A n c n ⃗x n
= λ n
m+1 c n ⃗x n
≈ λ n ⃗u m
S 2 S 3 ···S m S 2 S 3 ···S m
and so for large m, it must be the case that λ n ≈ S m+1 . This suggests the following procedure.
Procedure 7.88: Finding the Largest Eigenvalue with its Eigenvector
1. Start with a vector ⃗u 1 which you hope has a component in the direction of ⃗x n . The vector
(1,···,1) T is usually a pretty good choice.
2. If ⃗u k is known,
⃗u k+1 = A⃗u k
S k+1
where S k+1 is the entry of A⃗u k which has largest absolute value.
3. When the scaling factors, S k are not changing much, S k+1 will be close to the eigenvalue and
⃗u k+1 will be close to an eigenvector.
4. Check your answer to see if it worked well.
The shifted inverse power method involves finding the eigenvalue closest to a given complex number
along with the associated eigenvalue. If μ is a complex number and you want to find λ which is closest to
μ, you could consider the eigenvalues and eigenvectors of (A − μI) −1 .ThenA⃗x = λ⃗x if and only if
(A − μI)⃗x =(λ − μ)⃗x
If and only if
1
λ − μ ⃗x =(A − μI)−1 ⃗x
Thus, if λ is the closest eigenvalue of A to μ then out of all eigenvalues of (A − μI) −1 , the eigenvaue
given by 1
λ−μ would be the largest. Thus all you have to do is apply the power method to (A − μI)−1 and
the eigenvector you get will be the eigenvector which corresponds to λ where λ is the closest to μ of all
eigenvalues of A. You could use the eigenvector to determine this directly.