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According to this page: https://mathworld.wolfram.com/FrobeniusNorm.html the frobenius norm is equal to: $$\left\|A\right\|_F=\sqrt{\sum_{i=1}^m\sum_{i=1}^n |a_{i, j}|^2}$$ and I find it weird that it takes the absolute value on top of squaring it. Why is that? Wouldn't just squaring it suffice?

Sebastiano
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Mario
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    $A$ could be a complex matrix. That $|\cdot|$ is not just the absolute value, but also the modulus. – player3236 Nov 07 '20 at 12:27
  • @player3236 oh ok thanks. it was a dumb question now that I think about about it. I'm new in the matrix math field – Mario Nov 07 '20 at 12:29
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    @player3236 is correct. Notice that later on that same Mathworld page, it talks about the "conjugate transpose" of the matrix: another clue that it is a complex matrix. However, wouldn't it be best if that page said "complex matrix" right at the start? – GEdgar Nov 07 '20 at 12:30
  • @GEdgar but it could also be a matrix with only real elements right? – Mario Nov 07 '20 at 12:34
  • The conjugate transpose is the regular transpose, but all complex numbers are replaced with their conjugate. The use of "conjugate transpose" implies that complex matrices are considered. – player3236 Nov 07 '20 at 12:46

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