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Let $V$ be an inner product space generated over $\mathbb{C}$ and $B$ is a $n\times n$ normal complex matrix.

(1)I need to show that there exists a matrix $C$ such that $C^{2}=B$. I know that B is orthogonally diagonalizable by a theorem which was proved in class. Could I should that $C=BC^{-1}$?

(2) If the eigenvalues of $B$ are real, then $B$ is self-adjoint. Not sure where to start on this one.

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    Can you find a square root of a diagonal matrix with complex entries?2012-11-30

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Hint: Notice that if you have a matrix of the form $B=PDP^{-1}$ then $$B^2 = (PDP^{-1})(PDP^{-1}) = PD^2P^{-1}$$ Can you think of a square root for a diagonal matrix?

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    Wouldn't it just be the square roots of the eigenvalues of the eigenvectors that make up the basis for the complex vector space?2012-11-30
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    Yes, it would just be the square roots of the eigenvalues.2012-11-30
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    How could I show that if the eigenvalues are real, then B is self-adjoint? Is that because the matrix is orthogonally diagonalizable?2012-11-30
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    For every diagonal matrix, $D^T = D$, now you can use the fact that the eigenvalues are real.2012-11-30
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    I'm not following...2012-11-30
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    NVM, got it. thank you all!2012-11-30
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    I think Stefan meant $D^*$ instead of $D^{\mathrm{T}}$. A Hermitian matrix is always unitarily diagonalizable. The diagonal matrix then satisfies $D = D^*$. This means that for each eigenvalue, we necessarily have $\lambda = \overline{\lambda}$. This happens if and only if $\lambda$ is real.2012-11-30
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    How would that show that B is self-adjoint? would it be because $D^*D=D^*D$?2012-11-30
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    $$A^* = (P^*DP)^* = P^*D^*P = P^*DP=A$$2012-11-30
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Hint: $(ACA^{-1})^2=AC^2A^{-1}$