# how to prove $\|(A^HA)^k\| =||A||^{2k}$ using singular value decomposition

how to prove $\|(A^HA)^k\| =||A||^{2k}$ using singular value decomosition. $A^H$ is a hermitian matrix. $A$ element of $C^{p\times q}$, for every positive integer $k$.

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Presumably you are using the spectral norm, but for questions like this, one should specify what norm is used because there are many kinds of matrix norms. –  user1551 Jan 4 '13 at 13:55
yes, it is spectral norm –  aneshps Jan 4 '13 at 14:01
Recall that the spectral norm of a complex matrix $X$ is the largest sigular value of $X$. Let $A=U\Sigma V^\ast$ be a SVD of $A$, where $\Sigma=\operatorname{diag}(\sigma_1,\ldots,\sigma_n)$ are the (nonnegative) singular values of $A$ arranged in descending order. Then $\|A\|^{2k}=\sigma_1^{2k}$ and $\|(A^HA)^k\|=\|V\Sigma^{2k}V^\ast\|$. But $V\Sigma^{2k}V^\ast$ is already in the form of a SVD and its largest singular value is $\sigma_1^{2k}$. Hence the result.