# Show that norm of matrix A is given by the square root of the largest eigenvalue of A^tA

The norm is defined as ||A||=sup{ ||A*v*||} ||v||=1 I do not know why it is an eigenvalue of a product of A^tA not simply an eigenvalue of A. How to proceed? Thank You in advance for the answers.

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You write $^*$, then transpose: are you working with real of complex matrices? – Davide Giraudo Oct 14 '12 at 14:54
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. – kjetil b halvorsen Oct 14 '12 at 17:09

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, ..., 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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