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The norm is defined as $\|A\|=\sup\{ \|A v \| : \|v\|=1\}$. I want to show it is equal to the square root of the largest eigenvalue of $A^tA$.

I do not know why it is an eigenvalue of a product of $A^tA$ not simply an eigenvalue of $A$. How to proceed?

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    Are you sure your definition is written as you wanted? First, why use $A^*A$ and not simply $A$? Because $A$ might have complex eigenvalues! Second, you want to write $\| A \|=sup\{\| A^*A v \colon \| v \|=1 \}$ and in this case this is a well-known variational characterization of largest eigenvalue.2012-10-14

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The singular value decomposition of $A$ gives us orthogonal matrices $U, V$ and a diagonal matrix $S$ such that

$A = U S V^T$

Since $U$ and $ V$ are orthogonal, we have $\|U\| = \|V\| = 1$. Therefore, for all vectors $x$, we have

$\|A x\| = \|S x\|$

Since $S$ is the diagonal matrix containing the singular values of $A$ (which by definition are the roots of the eigenvalues of $A^T A$), the $x'$ which maximizes $\|S x\|$ is the unit vector $e_1 = (1, 0, \dots, 0)$ assuming the singular values in $S$ are sorted in descending order of magnitude.

Let $s$ be the largest singular value, then we have

$\|A\| = \|A x'\| = \|S x'\| = \|S e_1\| = \|e_1 s\| = s$

So, the norm of $A$ is indeed the largest singular value $s$ of $A$, which is the root of the largest eigenvalue of $A^T A$.

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    The vector needs to have norm 1, which is not the case for (1, 1, ..., 1)2017-11-03