Show $\Vert A\Vert_2 = \sup_{x \neq 0} \frac{x^T A x}{x^T x}$ where $A$ is symmetric and positive-definite

Question

Problem

Show: $$\Vert A\Vert_2 = \sup_{0 \neq x \in \mathbb{R}} \frac{x^T A x}{x^T x}$$ where $A$ is symmetric and positive definite.

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Try

Since

\begin{align} \Vert A\Vert_2 &= \sup_{0 \neq x \in \mathbb{R}} \frac{\Vert A x\Vert_2}{\Vert x\Vert_2} \\ &= \sup_{0 \neq x \in \mathbb{R}} \frac{x^T A^T A x}{x^T x} \end{align}

So I think the problem boils down to showing

$$ \sup_{x\neq0} x^T A x = \sup_{x\neq0} x^T A^T A x $$

where I'm stuck.

Any help will be appreciated.

Answer 1

Try showing that both sides are equal to the maximal eigenvalue of A, using the fact that there exists an orthonormal basis of eigenvectors.