14.1.4 Rayleigh Quotients And Estimates for Eigenvalues
There are many specialized results concerning the eigenvalues and eigenvectors for Hermitian
matrices. Recall a matrix A is Hermitian if A = A∗ where A∗ means to take the transpose of the
conjugate of A. In the case of a real matrix, Hermitian reduces to symmetric. Recall also that for
x ∈ Fn,
Recall the following corollary found on Page 475 which is stated here for convenience.
Corollary 14.1.8 If A is Hermitian, then all the eigenvalues of A are real and there exists
an orthonormal basis of eigenvectors.
this orthonormal basis,
For x ∈ Fn, x≠0, the Rayleigh quotient is defined by
Now let the eigenvalues of A be λ1 ≤ λ2 ≤
above orthonormal basis of eigenvectors mentioned in the corollary. Then if x
is an arbitrary
vector, there exist constants, ai
In other words, the Rayleigh quotient is always between the largest and the smallest eigenvalues of
A. When x = xn, the Rayleigh quotient equals the largest eigenvalue and when x = x1 the
Rayleigh quotient equals the smallest eigenvalue. Suppose you calculate a Rayleigh quotient. How
close is it to some eigenvalue?
Theorem 14.1.9 Let x≠0 and form the Rayleigh quotient,
Then there exists an eigenvalue of A, denoted here by λq such that
Proof: Let x = ∑
is the orthonormal basis of eigenvectors.
Now pick the eigenvalue λq which is closest to q. Then
which implies 14.5. ■
Example 14.1.10 Consider the symmetric matrix A =
. Let x
How close is the Rayleigh quotient to some eigenvalue of A
? Find the eigenvector and
eigenvalue to several decimal places.
Everything is real and so there is no need to worry about taking conjugates. Therefore, the
Rayleigh quotient is
According to the above theorem, there is some eigenvalue of this matrix λq such that
Could you find this eigenvalue and associated eigenvector? Of course you could. This is what
the shifted inverse power method is all about.
In other words solve
and divide by the entry which is largest, 3.8707, to get
and divide by the largest entry, 2.9979 to get
and divide by the largest entry, 3.0454, to get
and divide by the largest entry, 3.0421 to get
You can see these scaling factors are not changing much. The predicted eigenvalue is then
How close is this?
You see that for practical purposes, this has found the eigenvalue and an eigenvector.