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.

Weyl's inequality about perturbation
Let $A$ be Hermitian on inner product space $V$  with dimension $n$, with spectrum ordered in descending order $\lambda_1 \geq ... \geq \lambda_n$. Note that these eigenvalues can be ordered, because they are real (as eigenvalues of Hermitian matrices).

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

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In jargon, it says that $$\lambda_k$$ is Lipschitz-continuous on the space of Hermitian matrices with operator norm.

Weyl's inequality between eigenvalues and singular values
Let $$A \in \mathbb{C}^{n \times n}$$ have singular values $$\sigma_1(A) \geq \cdots \geq \sigma_n(A) \geq 0$$ and eigenvalues ordered so that $$|\lambda_1(A)| \geq \cdots \geq |\lambda_n(A)|$$. Then


 * $$ |\lambda_1(A) \cdots \lambda_k(A)| \leq \sigma_1(A) \cdots \sigma_k(A)$$

For $$k = 1, \ldots, n$$, with equality for $$k=n$$.

Estimating perturbations of the spectrum
Assume that $$R$$ is small in the sense that its spectral norm satisfies $$\|R\|_2 \le \epsilon$$ for some small $$\epsilon>0$$. Then it follows that all the eigenvalues of $$R$$ are bounded in absolute value by $$\epsilon$$. Applying Weyl's inequality, it follows that the spectra of the Hermitian matrices M and N are close in the sense that


 * $$|\mu_i - \nu_i| \le \epsilon \qquad \forall i=1,\ldots,n.$$

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 $$t>0$$ be arbitrarily small, and consider
 * $$M = \begin{bmatrix} 0 & 0 \\ 1/t^2 & 0 \end{bmatrix}, \qquad N = M + R = \begin{bmatrix} 0 & 1 \\ 1/t^2 & 0 \end{bmatrix}, \qquad R = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}.$$

whose eigenvalues $$\mu_1 = \mu_2 = 0$$ and $$\nu_1 = +1/t, \nu_2 = -1/t$$ do not satisfy $$|\mu_i - \nu_i| \le \|R\|_2 = 1$$.

Weyl's inequality for singular values
Let $$M$$ be a $$p \times n$$ matrix with $$1 \le p \le n$$. Its singular values $$\sigma_k(M)$$ are the $$p$$ positive eigenvalues of the $$(p+n) \times (p+n)$$ Hermitian augmented matrix


 * $$\begin{bmatrix} 0 & M \\ M^* & 0 \end{bmatrix}.$$

Therefore, Weyl's eigenvalue perturbation inequality for Hermitian matrices extends naturally to perturbation of singular values. This result gives the bound for the perturbation in the singular values of a matrix $$M$$ due to an additive perturbation $$\Delta$$:
 * $$|\sigma_k(M+\Delta) - \sigma_k(M)| \le \sigma_1(\Delta)$$

where we note that the largest singular value $$\sigma_1(\Delta)$$ coincides with the spectral norm $$\|\Delta\|_2$$.