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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.

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  • $\begingroup$ For what it's worth, $x^T A^T A x = ||Ax||_{2}^{2}$; i.e., the norm squared. Similarly, $x^T x = ||x||_{2}^{2}$. $\endgroup$ – avs Apr 17 at 23:46
  • $\begingroup$ 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. $\endgroup$ – DiegoMath Apr 17 at 23:51
  • $\begingroup$ 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 $\mathbb R^{m\times n}$? $\endgroup$ – Clayton Apr 17 at 23:54
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Try showing that both sides are equal to the maximal eigenvalue of A, using the fact that there exists an orthonormal basis of eigenvectors.

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