# Singular value decomposition inequality [closed]

Assume $A$ is nonsingular with SVD i.e $$A = U \Sigma V^{T}$$

$U,V$ orthogonal.

$\Sigma$ is a diagonal matrix with non-negative entries $\Sigma= diag (\sigma_{1}, \sigma_{2}, . . . , \sigma_{r})$ and $\sigma_1 ≥ \sigma_2 ≥\ldots ≥ \sigma_r> 0$ are the positive singular values of $A$.

Prove that

$\sigma_{n} \| x \| _{2} \leq \|Ax\|_{2} ≤ \sigma_{1}\|x\|_{2}$

Thanks!

## closed as off-topic by user21820, user223391, Did, Simply Beautiful Art, NamasteJan 14 '18 at 18:32

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You need to notice that, since $U$ and $V$ are orthogonal, $$\|Ax\|_2=\|\Sigma (V^Tx)\|_2\ \ \ \text {and } \ \ \|V^Tx\|_2=\|x\|_2$$
• That should be $\|Ax\| = \|\Sigma(V^Tx)\|$, and of course $\|V^Tx\| = \|x\|$. – Omnomnomnom Jul 6 '17 at 13:57
• Yes, thanks.  – Martin Argerami Jul 6 '17 at 14:03