Largest eigenvalue of a real symmetric matrix

Question

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.

Answer 1

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!

Answer 2

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)

Answer 3

Another hint along the same lines as Matt's: for which $\vec{v}$ is the LHS of your inequality maximised?