The Courant-Fischer theorem states that $$\lambda_j=\max_{\dim(\mathbb{V})=j}\min_{v\in \mathbb{V},v\neq 0}\rho(v,A)=\min_{\dim(\mathbb{W})=n-j+1}\max_{w\in \mathbb{W},w\neq0}\rho(v,A)$$ where $\lambda_j$ is the $j$th entry of the largest to smallest sequence of eigenvalues of a Hermitian matrix $A$. $\rho(v,A)$ denotes the Rayleigh quotient.

We must show Weyl’s inequality:

Let $B=A+E$ where $A$ and $E$ are hermitian matrices and $\lambda_j$ and $\mu_j$ denote the $j$th eigenvalue (largest to smallest) of $B$ and $A$ respectively. Then $$|\lambda_j-\mu_j|\leq ||E||_2$$

I tried squaring both sides so that I could use Courant-Fischer on the largest eigenvalue of $E$ but I wound up with very complicated expressions on both sides of the inequality.

I’ve tried looking for references online, but every document I find states a version of Weyl’s inequality that is so different I would need to make another MSE post entirely to ask for a proof of equivalency.


1 Answer 1


Since $E$ is Hermitian, it is unitarily diagonalisable. Hence $|\rho(w,E)|\le\|E\|_2$ for every $w\ne0$ and \begin{aligned} \lambda_j &=\min_{\dim W=n-j+1}\max_{w\in W\setminus0}\rho(w,A)\\ &=\min_{\dim W=n-j+1}\left(\max_{w\in W\setminus0}\rho(w,B)+\rho(w,E)\right)\\ &\le\min_{\dim W=n-j+1}\left(\max_{w\in W\setminus0}\rho(w,B)+\|E\|_2\right)\\ &=\mu_j+\|E\|_2.\end{aligned} Similarly, by interchanging the roles of $A$ and $B$ in the above, we also have $\mu_j\le\lambda_j+\|E\|_2$. Combining the two inequalities, we get $|\lambda_j-\mu_j|\le\|E\|_2$.

  • $\begingroup$ How did you get $|\rho(w,E)|\leq ||E||_2$? I thought you could only safely say that $\rho(w,E)$ is bounded above by the largest eigenvalue of $E$, not the largest singular value. $\endgroup$
    – Ryan
    Feb 24, 2020 at 17:19
  • $\begingroup$ @RyanGreyling The spectral radius of a Hermitian matrix is identical to the largest singular value. That's why the induced $2$-norm is sometimes (misleadingly) called the spectral norm. $\endgroup$
    – user1551
    Feb 24, 2020 at 18:03
  • $\begingroup$ Why is that true? $\endgroup$
    – Ryan
    Feb 24, 2020 at 18:19
  • $\begingroup$ I figured out why. Hermitian matrices can be decomposed to $U^*\Lambda U$ because its left and right singular vectors are the same. $U$ is unitary so it preserves vector norm. Thus we see for Hermitian matrices that the Rayleigh quotient can be seen as a weighted average of the singular values, which is bounded above by the largest singular value. $\endgroup$
    – Ryan
    Feb 25, 2020 at 18:26
  • 1
    $\begingroup$ @RyanGreyling For each eigenvalues $\lambda_i$ of $U\Lambda U^\ast$, let $\lambda_i=|\lambda_i|w_i$, where $w_i$ is a complex number lying on the unit circle. Let $W=\operatorname{diag}(w_1,\ldots,w_n)$ and $|\Lambda|=\operatorname{diag}(|\lambda_1|,\ldots,|\lambda_n|)$. Then $U\Lambda U^\ast=U|\Lambda|(WU^\ast)=U|\Lambda|(UW^\ast)^\ast$, but the RHS is a singular value decomposition, as $U$ and $UW^\ast$ are unitary and $|\Lambda|$ is a nonnegative diagonal matrix. Hence the singular values of $U\Lambda U^\ast$ are identical to the magnitudes of the eigenvalues. $\endgroup$
    – user1551
    Feb 25, 2020 at 18:37

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