# Showing there exists a matrix $C^{2}=B$

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? – alex.jordan Nov 30 '12 at 20:25

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? – MathScratch Nov 30 '12 at 20:32
Yes, it would just be the square roots of the eigenvalues. – EuYu Nov 30 '12 at 20:36
How could I show that if the eigenvalues are real, then B is self-adjoint? Is that because the matrix is orthogonally diagonalizable? – MathScratch Nov 30 '12 at 20:40
For every diagonal matrix, $D^T = D$, now you can use the fact that the eigenvalues are real. – Stefan Nov 30 '12 at 20:42
I'm not following... – MathScratch Nov 30 '12 at 20:50

Hint: $(ACA^{-1})^2=AC^2A^{-1}$

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