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

### 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.

## 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.

• For what it's worth, $x^T A^T A x = ||Ax||_{2}^{2}$; i.e., the norm squared. Similarly, $x^T x = ||x||_{2}^{2}$. – avs Apr 17 at 23:46
• And what if $A=kI$, where $k$ is a positive real number and $I$ is the identity matrix? In this case, $A$ is a symmetric positive definite matrix, but the norms are diferents, in fact we obtain $k$ and $k^2$ in each of ones. – DiegoMath Apr 17 at 23:51
• I edited your post; please ensure that it still remains what you wanted. Also, do you really mean $x\in\mathbb R$? Or do you maybe mean $x\in\mathbb R^n$ or $\mathbb R^{m\times n}$? – Clayton Apr 17 at 23:54