In linear algebra, Weyl's inequality is a theorem about the changes to eigenvalues of an Hermitian matrix that is perturbed. It can be used to estimate the eigenvalues of a perturbed Hermitian matrix.
Let be Hermitian on inner product space with dimension , with spectrum ordered in descending order . Note that these eigenvalues can be ordered, because they are real (as eigenvalues of Hermitian matrices).
Weyl's inequality states that the spectrum of Hermitian matrices is stable under perturbation. Specifically, we have:
In jargon, it says that
λk
Let
A\inCn
\sigma1(A)\geq … \geq\sigman(A)\geq0
|λ1(A)|\geq … \geq|λn(A)|
|λ1(A) … λk(A)|\leq\sigma1(A) … \sigmak(A)
For
k=1,\ldots,n
k=n
Assume that
R
\|R\|2\le\epsilon
\epsilon>0
R
\epsilon
|\mui-\nui|\le\epsilon \foralli=1,\ldots,n.
Note, however, that this eigenvalue perturbation bound is generally false for non-Hermitian matrices (or more accurately, for non-normal matrices). For a counterexample, let
t>0
M=\begin{bmatrix}0&0\ 1/t2&0\end{bmatrix}, N=M+R=\begin{bmatrix}0&1\ 1/t2&0\end{bmatrix}, R=\begin{bmatrix}0&1\ 0&0\end{bmatrix}.
\mu1=\mu2=0
\nu1=+1/t,\nu2=-1/t
|\mui-\nui|\le\|R\|2=1
Let
M
p x n
1\lep\len
\sigmak(M)
p
(p+n) x (p+n)
\begin{bmatrix}0&M\ M*&0\end{bmatrix}.
Therefore, Weyl's eigenvalue perturbation inequality for Hermitian matrices extends naturally to perturbation of singular values.[3] This result gives the bound for the perturbation in the singular values of a matrix
M
\Delta
|\sigmak(M+\Delta)-\sigmak(M)|\le\sigma1(\Delta)
\sigma1(\Delta)
\|\Delta\|2