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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##.
A Hermitian matrix is a square matrix that is equal to its own conjugate transpose. In other words, the elements above the main diagonal are the complex conjugates of the elements below the main diagonal.
To prove this, we can use the definition of a Hermitian matrix and the properties of matrix multiplication. First, we can write out the left side of the equation as ##H^2 = HH##. Then, we can use the definition of a Hermitian matrix to rewrite the right side as ##H^\dagger H = H^*H##. Finally, we can use the properties of matrix multiplication to show that ##HH = H^*H##, proving that ##H^2 = H^\dagger H##.
This property is important because it allows us to simplify calculations and proofs involving Hermitian matrices. It also helps us to better understand the properties and behavior of these matrices.
One example of a Hermitian matrix that satisfies this property is the identity matrix, which is equal to its own conjugate transpose and also its own square.
The eigenvalues of a Hermitian matrix are always real numbers. This property can be proven using the fact that ##H^2 = H^\dagger H##, which shows that the eigenvalues of ##H^2## are the squares of the eigenvalues of ##H##. Since the eigenvalues of ##H^2## are real, this means that the eigenvalues of ##H## must also be real.