Weyl's Inequality

Hermitian eigenvalues are stable: | _k(A+B) - _k(A)| ≤ |B|_ op.

The idea

Theorem (Weyl's inequality (perturbation form)).

Let $A$ and $B$ be $n \times n$ Hermitian matrices, and for each $k$ let $\lambda_k(\cdot)$ denote the $k$-th eigenvalue in sorted order. Then $|\lambda_k(A + B) - \lambda_k(A)| \leq \|B\|_{\text{op}}, \quad k = 1, \ldots, n,$ where the operator norm $\|B\|_{\text{op}}$ is the largest factor by which $B$ stretches any unit vector.

The inequality answers a stability question: how far can the eigenvalues of a matrix move when the matrix is disturbed slightly, by measurement error, rounding, or any small perturbation $B$? Eigenvalues are roots of the characteristic polynomial, and the roots of a polynomial can move a long way in response to a small change in its coefficients, so nothing guarantees stability in general.

For Hermitian matrices, Weyl's inequality guarantees it. Perturbing $A$ by $B$ moves each eigenvalue, matched by index, by no more than the size of the perturbation itself. Every eigenvalue is bounded individually and by the same amount, so no single one can absorb the whole disturbance.

The Hermitian hypothesis cannot be dropped: the eigenvalues of a general matrix can be far more sensitive than the perturbation is large.

Ways to work on it

• Walkthrough. Verify Weyl's perturbation bound on a concrete 2 × 2 example.
• Practice. Perturb a diagonal matrix and check its eigenvalue shifts against Weyl's bound.
• Hardest. Apply the general form of Weyl's inequality and derive the perturbation bound from it.

Not sure where to start? Take the ten-question placement test.