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EDITED after found false

Define $|v|$ as the normal Euclidean norm. $A$ is a n by n matrix.

Define the spectral norm: $$|A| = \max \frac{|Ax|}{|x|}$$

I need to proove that: $|A|$ is the maximum of $\sqrt{\lambda}$ such that $A^TA-\lambda I$ is singular, without using the eigenvalues, eigenvectors or diagonalization.

I saw that if $A^TA-\lambda I$ is singular then there is $x$ such that $A^TAx=\lambda x$ then $|A| \geq \sqrt{\lambda} $, but I can see the other way.

Can someone help?

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What you are proving is one of the theorems that leads up to the singular value decomposition. It may help you to look that up. –  Ian Jul 6 at 18:41
    
I want to have a simple proof of this particular problem for a particular purpose, I will look up about the decomposition, thanks. –  Mr.T Jul 6 at 18:44

1 Answer 1

It's not true. For example, try $A = \pmatrix{0 & 1\cr 0 & 0\cr}$.

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I will look again at the problem and fix it. –  Mr.T Jul 6 at 18:32
    
I have fixed the problem. Thanks for the example. –  Mr.T Jul 6 at 18:39

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