Weyl's inequality

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 ${\displaystyle M=N+R,\,N,}$ and ${\displaystyle R}$ be n×n Hermitian matrices, with their respective eigenvalues ${\displaystyle \mu _{i},\,\nu _{i},\,\rho _{i}}$ ordered as follows:

${\displaystyle M:\quad \mu _{1}\geq \cdots \geq \mu _{n},}$

${\displaystyle N:\quad \nu _{1}\geq \cdots \geq \nu _{n},}$

${\displaystyle R:\quad \rho _{1}\geq \cdots \geq \rho _{n}.}$

Then the following inequalities hold:

${\displaystyle \nu _{i}+\rho _{n}\leq \mu _{i}\leq \nu _{i}+\rho _{1},\quad i=1,\dots ,n,}$

and, more generally,

${\displaystyle \nu _{j}+\rho _{k}\leq \mu _{i}\leq \nu _{r}+\rho _{s},\quad j+k-n\geq i\geq r+s-1.}$

In particular, if ${\displaystyle R}$ is positive definite then plugging ${\displaystyle \rho _{n}>0}$ into the above inequalities leads to

${\displaystyle \mu _{i}>\nu _{i}\quad \forall i=1,\dots ,n.}$

Note that these eigenvalues can be ordered, because they are real (as eigenvalues of Hermitian matrices).

Let ${\displaystyle A\in \mathbb {C} ^{n\times n}}$ have singular values ${\displaystyle \sigma _{1}(A)\geq \cdots \geq \sigma _{n}(A)\geq 0}$ and eigenvalues ordered so that ${\displaystyle |\lambda _{1}(A)|\geq \cdots \geq |\lambda _{n}(A)|}$. Then

${\displaystyle |\lambda _{1}(A)\cdots \lambda _{k}(A)|\leq \sigma _{1}(A)\cdots \sigma _{k}(A)}$

For ${\displaystyle k=1,\ldots ,n}$, with equality for ${\displaystyle k=n}$. [1]

Assume that we have a bound on R in the sense that we know that its spectral norm (or, indeed, any consistent matrix norm) satisfies ${\displaystyle \|R\|_{2}\leq \epsilon }$. Then it follows that all its eigenvalues are bounded in absolute value by ${\displaystyle \epsilon }$. Applying Weyl's inequality, it follows that the spectra of M and N are close in the sense that[2]

${\displaystyle |\mu _{i}-\nu _{i}|\leq \epsilon \qquad \forall i=1,\ldots ,n.}$

The singular values {σ_k} of a square matrix M are the square roots of eigenvalues of M*M (equivalently MM*). Since Hermitian matrices follow Weyl's inequality, if we take any matrix A then its singular values will be the square root of the eigenvalues of B=A*A which is a Hermitian matrix. Now since Weyl's inequality hold for B, therefore for the singular values of A.[3]

This result gives the bound for the perturbation in the singular values of a matrix A due to perturbation in A.