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.
Best Answer
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 $|\lambda_j-\mu_j|\le\|E\|_2$.