Weyl's inequality for largest eigenvalues of two real symmetric matrices I am trying to show that
$| \lambda_n(A) - \lambda_n(B)| \leq \lVert{\mathbf{A} - \mathbf{B}} \rVert_F $
where $A, B \in \mathbb{R}_{sym}^{nxn}$ and $\lambda_n(A)$ denotes the largest eigenvalue of $A$.
The largest eigenvalue for $A$ is defined as $\max_{v \in \mathbb{R}^n,   \lVert v \rVert_2 = 1}  v^TAv$, and for B, is similar.
I know this is related to Weyl's Inequality in some sense but cannot construct proof to show this.
 A: If we use the property you have it says
"The largest eigenavalue for A is defined as $ \max_{v \in \mathbb{R} ,  \|v\|=1} v^{T}Av$ "
Now if $A,B$ are symmetric then $A-B$ is also symmetric and using that we have
$$ \lambda_{\text{max}}(A-B) = \max_{v \in \mathbb{R} ,  \|v\|=1} v^{T}(A-B)v $$
Now you have to show the rest using properties of $\max$ note that
$$ \max\{ f(x) + g(x) \} \leq \max \{ f(x)\} + \max \{ g(x) \}$$
and if $\theta < 0$
$$ \max \theta f(x) = \theta \min f(x)  $$
I think this may help.
A: With an additional assumption, you don't even need to argue with $\operatorname{max}$.
Suppose that A and B are positively semidefinite. We have [1, 2], $A=M^2, \ B=N^2$ for some $M,N$ symmetric. $\Vert{A-B}\Vert_F \\ = \Vert M^2-N^2\Vert_F \\ \geq \Vert M^2-N^2\Vert_2 \\ \geq \left| \Vert M^2 \Vert_2 - \Vert N^2\Vert_2 \right| \\
= \left| \Vert M \Vert_2^2-\Vert N \Vert_2^2\right| \\
= \left| \sqrt{\lambda_{max}(M^2)}^2 - \sqrt{\lambda_{max}(N^2)}^2\right| \\
= \left |\lambda_{max}(A) - \lambda_{max}(B)\right|$.
