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Euge
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Show that an ##n\times n##-matrix ##H## is hermitian if and only if ##H^2 = H^\dagger H##.
I get:Infrared said:@julian I didn't fully read your posts, but it looks like you're making a bit of a mountain out of a molehill.
If ##H## is Hermitian, meaning ##H=H^\dagger,## then multiply both sides by ##H## on the right to get ##H^2=H^\dagger H.##
On the other hand, if ##H^2=H^\dagger H##, then to show that ##H=H^\dagger##, you just have to check that ##||Hx-H^\dagger x||^2=\langle Hx-H^\dagger x,Hx-H^\dagger x\rangle## is always zero, which follows quickly from expanding the inner product and using the given relation.
Edit: Spoiler tags
You use this in the proof of the spectral theorem for Hermitian matrices.Infrared said:If you already know that ##H## was Hermitian, you could say ##\langle H x,v \rangle =\langle x, Hv \rangle=0##
But this still isn't rightStructure seeker said:You can use ## {H^\dagger}^2 V=H^\dagger H V=0## to prove that the kernel ##V## of ##H## is also that of ##H^\dagger## since the rank of ##H## coincides with the rank of ##H^\dagger##. That step should be inserted for arguing that ##v## is also a zero-eigenvalue eigenvector of ##H^\dagger##.