# Linear algebra eigenvalues and diagonalizable matrix

Let $A$ be an $n\times n$ matrix over $\mathbb{C}$. First I don't understand why $AA^*$ can be diagnosable over $\mathbb{C}$. And why $i+1$ can't be eigenvalue of $AA^*$?

Hope question is clear enough and I don't have any spelling mistake and used right expressions.

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Do you know that any hermitian matrix is diagonalizable in $\mathbb{C}$? What do you mean with $I$ in $I+1$? –  Git Gud Feb 2 '13 at 22:46
i+1 as I from complex numbers –  Mary Feb 2 '13 at 22:49
For your second question, assume $AA^*X=(1+i)X$ for some nonzero eigenvector $X$ (so that $X^*X>0$). Then $X^*AA^*X=(1+i)X^*X$ so $1+i=(A^*X)^*A^*X/X^*X$. Now observe that for every vector $Y$, $Y^*Y$ is a nonegative number. This is how you can prove in general that the spectrum of $AA^*$ is contained in $[0,+\infty)$. –  1015 Feb 2 '13 at 22:59
@Mary Next time try to provide a little context so people know what you can use. –  Git Gud Feb 2 '13 at 23:03

$\textbf{Hint:}$ Any hermitian matrix is diagonalizable. Prove, by definition, that $AA^*$ is a hermitian.

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Thank you easy to prove that now but still can't understand the second:\ –  Mary Feb 2 '13 at 22:51
@Mary Do you know that any hermitian matrix is normal and any normal matrix is unitarily similar to a diagonal matrix? –  Git Gud Feb 2 '13 at 22:54

Any Hermitian matrix is diagonalizable. All eigenvalues of a Hermitian matrix are real. These two facts (that you probably learnt) solve the question: Show your matrix is Hermitian and note that $1+i$ is not real.

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