Prove spectral norm $\|A\|\geq x^T A x$, $\forall x$ where $\|x\|_2=1$

Question

I'm not sure if this is true, as it is something I am inferring from a proof that I am trying to understand.

I know that $\|A\|=\max\{-\lambda_{\min}(A),\lambda_\max(A)\}$ for $A$ symmetric. However I don't know how to prove the statement in the title.

Answer 1

If $A$ is symmetric it’s diagonalizable and has an orthonormal basis $v_i$ with eigenvalues $\lambda_i$.

Then write $x=\sum_i (x,v_i)v_i$, so that:

$$|x^TAx|=|\sum_{i=1} (x,v_i)^2\lambda_i|\leq |\lambda_M| \|x\|^2,$$

Can you finish from here?

Answer 2

Because $A$ is symmetric, there exists an orthonormal basis of eigenvectors. You should decompose $x$ in such a basis to prove the statement.

Answer 3

Posting this as a separate answer so the solution is complete.

Let $A=\sum_i\lambda_iv_iv_i^T$, where $\{v_i\}_i$ is an orthonormal eigenbasis of $A$, as $A$ is symmetric. Then,

$$x^TAx=x^TA\sum_i v_i v_i^Tx = x^T\sum_i \lambda_iv_i v_i^T x \leq x^T \max_i\{\lambda_i\}x=\max_i\{\lambda_i\}=\|A\|.$$