How to prove, for any vector $v$ and matrix $A$, $\langle Av,v\rangle\le \lambda\langle v,v\rangle$, where $\lambda$ is the maximum eigenvalue of $A$

Question

As in the title, let $A$ be any real matrix, how to prove that, for any vector $v$, we have $$ \langle Av,v\rangle \leq \lambda_+(A)\langle v,v\rangle,$$ where $\lambda_+(A)$ is the maximum eigenvalue of $A$?

Answer 1

This is not true. It's true if $A$ is diagonalizable. For a counterexample to the question as stated let $A=\begin{bmatrix}0&1\\0&0\end{bmatrix}$, $v=\begin{bmatrix}1\\1\end{bmatrix}$; then $\langle v,Av\rangle=1$ while $\lambda_+(A)=0$.

Answer 2

This can fail in several ways:

1. as David Ullrich mentioned, it can fail if $A$ is not diagonalizable;
2. it may fail if $A$ does not have real eigenvalues (the inequality may not even make sense in this case);
3. it can fail if $A$ is not unitarily diagonalizable.

To see the third point, let $$A=\begin{bmatrix} 1&10 \\ 0&1\end{bmatrix}\begin{bmatrix} 1&0\\ 0&2\end{bmatrix}\begin{bmatrix} 1&-10\\0&1 \end{bmatrix}=\begin{bmatrix}1& 10\\0& 2\end{bmatrix}.$$

Then for $v=[1,1]^{T},$ we get $\langle Av,v\rangle=13,$ but $2\langle v,v\rangle=2(2)=4$.