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 ${\textstyle A}$ be Hermitian on inner product space ${\textstyle V}$ with dimension ${\textstyle n}$, with spectrum ordered in descending order ${\textstyle \lambda _{1}\geq ...\geq \lambda _{n}}$. Note that these eigenvalues can be ordered, because they are real (as eigenvalues of Hermitian matrices).^[1]

Weyl inequality — 

$${\displaystyle \lambda _{i+j-1}(A+B)\leq \lambda _{i}(A)+\lambda _{j}(B)\leq \lambda _{i+j-n}(A+B)}$$

Proof

By the min-max theorem, it suffices to show that any ${\textstyle W\subset V}$ with dimension ${\textstyle i+j-1}$, there exists a unit vector ${\textstyle w}$ such that ${\textstyle \langle w,(A+B)w\rangle \leq \lambda _{i}(A)+\lambda _{j}(B)}$.

By the min-max principle, there exists some ${\textstyle W_{A}}$ with codimension ${\textstyle (i-1)}$, such that

$${\displaystyle \lambda _{i}(A)=\max _{x\in W_{A};\|x\|=1}\langle x,Ax\rangle }$$

Similarly, there exists such a ${\textstyle W_{B}}$ with codimension ${\textstyle j-1}$. Now ${\textstyle W_{A}\cap W_{B}}$ has codimension ${\textstyle \leq i+j-2}$, so it has nontrivial intersection with ${\textstyle W}$. Let ${\textstyle w\in W\cap W_{A}\cap W_{B}}$, and we have the desired vector.

The second one is a corollary of the first, by taking the negative.

Weyl's inequality states that the spectrum of Hermitian matrices is stable under perturbation. Specifically, we have:^[1]

Corollary (Spectral stability) — 

$${\displaystyle \lambda _{k}(A+B)-\lambda _{k}(A)\in [\lambda _{n}(B),\lambda _{1}(B)]}$$

$${\displaystyle |\lambda _{k}(A+B)-\lambda _{k}(A)|\leq \|B\|_{op}}$$

where

$${\displaystyle \|B\|_{op}=\max(|\lambda _{1}(B)|,|\lambda _{n}(B)|)}$$

is the operator norm.

In jargon, it says that ${\displaystyle \lambda _{k}}$ is Lipschitz-continuous on the space of Hermitian matrices with operator norm.

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 ${\displaystyle k=n}$. ^[2]