Frobenius norm of the difference of two symmetric matrices.

Question

Let A and B be the $n\times n$ symmetric matrices. Let $\lambda_A, \lambda_B$ be the largest eigen values of A and B. Show that $$|\lambda_A-\lambda_B|\leq \|A-B\|_F.$$
Here $\|A\|_F$ is the Frobenius norm.
I want to use the fact that $$\lambda_A=\max_{\|v\|=1, v\in\mathbb{R}^n}v^TAv.$$ Please help how to do this.

Answer 1

Hint. Prove that $\lambda_\max(A)\le\lambda_\max(A-B)+\lambda_\max(B)\le\|A-B\|_F+\lambda_\max(B)$. It follows that $\lambda_\max(A)-\lambda_\max(B)\le\|A-B\|_F$ and by symmetry, also that $\lambda_\max(B)-\lambda_\max(A)\le\|B-A\|_F$. Hence the conclusion follows.