# Equality for matrix norm product of matrix * it's transpose, and square of the norm of the matrix [closed]

So, I know that $||AB|| \leq ||A||\cdot||B||$ (2-norm)

I'm doing a work on matrix algorithms and i seem to get as a result that $||A^TA|| = ||A||^2$

Does this always apply, or when and why does it happen?

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## closed as too localized by Martin Brandenburg, Suvrit, Matthew Daws, Igor Rivin, Denis SerreDec 13 '11 at 15:21

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Which matrix norm are you using? –  Igor Rivin Dec 13 '11 at 14:05
2-norm, sorry for not mentioning that –  Koeneuze Dec 13 '11 at 14:15
I think this would be much better asked on math.stackexchange.com as it's not really research level. –  Matthew Daws Dec 13 '11 at 14:53
...and, if you do ask at stackexchange, it would be better to give, explicitly, what your definition of the 2-norm is. –  Matthew Daws Dec 13 '11 at 14:59

If I'm not missing something, this is very obvious: $|A|^2_2$ is the maximal absolute value of the eigenvalues of $B = A^T A$. Thus $|A^TA|_2 =|B|_2$ is the square root of the maximal absolute value of the eigenvalues of $B^TB = B^2$. Since $B$ is diagonalizable, the eigenvalues of $B^2$ are just the squares of those of $B$.