Prove that the eigenvalues of a Hermitian matrix H must be real numbers.

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May 14th, 2015

Let A be a Hermitian matrix.

Then, by definition:

A=A

where A denotes the conjugate transpose of A.


Let λ be an eigenvalue of A.

Let v be an eigenvector corresponding to the eigenvalue λ.

By definition of eigenvector:

Avv

Left-multiplying both sides by v, we obtain:

(1):vAv=vλvvv


Firstly, note that both vAv and vv are 1×1-matrices.

Now observe that, using Conjugate Transpose of Matrix Product: General Case:

(vAv)=vA(v)

As A is Hermitian, and (v)=v by Double Conjugate Transpose is Itself, it follows that:

vA(v)=vAv

That is, vAv is also Hermitian.


By Product with Conjugate Transpose Matrix is Hermitian, vv is Hermitian.

So both vAv and vv are Hermitian 1×1 matrices.


Now suppose that we have for some a,b∈C:

vAv=[a]
vv=[b]

Note that b≠0 as an Definition:Eigenvector is non-zero.

By definition of a Hermitian matrix:

a=a¯ and b=b¯

where a¯ denotes the complex conjugate of a.

By Complex Number equals Conjugate iff Wholly Real, it follows that a,b∈R, that is, are real.


May 14th, 2015

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