Largest eigenvalue of a real symmetric matrix If $\lambda$ is the largest eigenvalue of a real symmetric $n \times n$ matrix $H$, how can I show that: $$\forall v \in \mathbb{R^n}, ||v||=1 \implies v^tHv\leq \lambda$$
Thank you.
 A: Hint: 
Real symmetric matrices are diagonalizable.
Hint 2 (added after reading comments on posts):
A matrix is diagonalizable by a suitable choice of coordinates if and only if there is an eigenbasis. (taken from here)
A: Step 1: All Real Symmetric Matrices can be diagonalized in the form: 
$
H = Q\Lambda Q^T
$
So,  $ {\bf v}^TH{\bf v} = {\bf v}^TQ\Lambda Q^T{\bf v} $
Step 2: Define transformed vector: $ {\bf y} = Q^T{\bf v} $. 
So, $ {\bf v}^TH{\bf v} = {\bf y}^T\Lambda {\bf y} $ 
Step 3: Expand  
$ {\bf y}^T\Lambda {\bf y}  = \lambda_{\max}y_1^2 + \lambda_{2}y_2^2 + \cdots + \lambda_{\min}y_N^2 $
\begin{eqnarray}
\lambda_{\max}y_1^2 + \lambda_{2}y_2^2 + \cdots + \lambda_{\min}y_N^2& \le & \lambda_{\max}y_1^2 + \lambda_{\max}y_2^2 + \cdots + \lambda_{\max}y_N^2 \\
 &  & =\lambda_{\max}(y_1^2 +y_2^2 + \cdots y_N^2) \\
 &  & =\lambda_{\max} {\bf y}^T{\bf y} \\
\implies {\bf y}^T\Lambda {\bf y} & \le & \lambda_{\max} {\bf y}^T{\bf y}
\end{eqnarray}
Step 5: Since $Q^{-1} = Q^T, QQ^T = I $
\begin{eqnarray}
{\bf y}^T{\bf y} &= &{\bf v}^TQQ^T{\bf v} = {\bf v}^T{\bf v}
\end{eqnarray}
Step 6: Putting it all back together 
\begin{eqnarray}
{\bf y}^T\Lambda {\bf y} & \le & \lambda_{\max} {\bf y}^T{\bf y} \\
{\bf v}^TH{\bf v}  & \le & \lambda_{\max}{\bf v}^T{\bf v}
\end{eqnarray}
By definition, $ {\bf v}^T{\bf v} = \|{\bf v}\|^2 $ and by definition $\|{\bf v}\| = 1$
\begin{eqnarray}
{\bf v}^TH{\bf v}  & \le & \lambda_{\max}
\end{eqnarray}
Boom! 
A: Another hint along the same lines as Matt's: for which $\vec{v}$ is the LHS of your inequality maximised?
