Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 1 down vote accepted

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

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.