Linear Algebra, Theory And Applications, 2012a

390 NUMERICAL METHODS FOR FINDING EIGENVALUES
A case of all this is of great interest. Suppose A has a largest eigenvalue λ which is
real. Then A n is of the form ( )
A n−1 a 1 , ··· ,A n−1 a n and so likely each of these columns
will be pointing roughly in the direction of an eigenvector of A which corresponds to this
eigenvalue. Then when you do the QR factorization of this, it follows from the fact that R
is upper triangular, that the first column of Q will be a multiple of A n−1 a 1 and so will end
up being roughly parallel to the eigenvector desired. Also this will require the entries below
the top in the first column of A n = Q T AQ will all be small because they will be of the form
q T i Aq 1 ≈ λq T i q 1 =0. Therefore, A n will be of the form
( )
λ
′
a
e B
where e is small. It follows that λ ′ will be close to λ and q 1 will be close to an eigenvector for
λ. Then if you like, you could do the same thing with the matrix B to obtain approximations
for the other eigenvalues. Finally, you could use the shifted inverse power method to get
more exact solutions.
15.2.2 The Case Of Real Eigenvalues
With these lemmas, it is possible to prove that for the QR algorithm and certain conditions,
the sequence A k converges pointwise to an upper triangular matrix having the eigenvalues
of A down the diagonal. I will assume all the matrices are real here. ( ) 0 1
This convergence won’t always happen. Consider for example the matrix
.
1 0
You can verify quickly that the algorithm will return this matrix for each k. The problem
here is that, although the matrix has the two eigenvalues −1, 1, they have the same absolute
value. The QR algorithm works in somewhat the same way as the power method, exploiting
differences in the size of the eigenvalues.
If A has all real eigenvalues and you are interested in finding these eigenvalues along
with the corresponding eigenvectors, you could always consider A + λI instead where λ is
sufficiently large and positive that A + λI has all positive eigenvalues. (Recall Gerschgorin’s
theorem.) Then if μ is an eigenvalue of A + λI with
(A + λI) x = μx
then
Ax =(μ − λ) x
so to find the eigenvalues of A you just subtract λ from the eigenvalues of A + λI. Thus
there is no loss of generality in assuming at the outset that the eigenvalues of A are all
positive. Here is the theorem. It involves a technical condition which will often hold. The
proof presented here follows [26] and is a special case of that presented in this reference.
Before giving the proof, note that the product of upper triangular matrices is upper
triangular. If they both have positive entries on the main diagonal so will the product.
Furthermore, the inverse of an upper triangular matrix is upper triangular. I will use these
simple facts without much comment whenever convenient.
Theorem 15.2.4 Let A be a real matrix having eigenvalues
λ 1 >λ 2 > ···>λ n > 0
and let
A = SDS −1 (15.11)