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Orthgonal matirx Q : Suppose that $A=Q\lambda Q^{T}$ iff $A=A^{T}$ (A is symmetric)
It means A's eigenvectors are orthogonal and unit length.

By the property of Hermitian martrix " If $A=A^{H}$, every eigenvalue is real"
the matirx $\lambda$ is real.
Of course, this is true because the matrix A has only real components.
(Note that real components are just $b=0$ in $a+ib$ form)

Now, let's consider the matirx A has complex components.

U is a unitary matrix : $A=U\lambda U^{H}$ and it means $A=A^{H}$? How can I prove this?
The book says, the matirx $\lambda$ is also real in here. But I'm not sure how do we know that.
If we conclude that $A=A^{H}$ from above, the matirx $\lambda$ must be real, that's right. The problem is that I have no idea how can I prove this...

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1 Answer 1

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If your diagonal matrix $D$ is real then it is true that $A^\rm{H} = A$. We have $$A^\mathrm{H} = (U^\mathrm{H}DU)^\mathrm{H}=U^\mathrm{H}D^\mathrm{H}U=U^\mathrm{H}DU=A$$ so that your matrix is Hermitian. If $D$ is not real then the most we can conclude is that $A$ is normal: $$A^\mathrm{H}A = AA^\mathrm{H}$$

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