# Proof of existence of square root of unitary and symmetric matrix

I'm struggling with this exercise

Let $U$ be a unitary and symmetric matrix ($U^T = U$ and $U^*U = I$).

Prove that there exists a complex matrix $S$ such that:

• $S^2 = U$
• $S$ is a unitary matrix
• $S$ is symmetric
• Each matrix that commutes with $U$, commutes with $S$
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Unitary symmetric? Are you sure? If the exercise regards symmetric definite positive matrices, you could try generalizing this ancient method for computing a square root: en.wikipedia.org/wiki/… –  Giuseppe Negro Jun 4 '12 at 17:33
@user: If you are still looking for some help with this, I could post something in a few hours. –  cardinal Jun 6 '12 at 2:21
I would be very grateful if you did, @cardinal –  user1111261 Jun 6 '12 at 16:03
@Yuki: Complex symmetric matrices are not necessarily diagonalizable. I have some patching to do. :) –  cardinal Jun 8 '12 at 2:03
@cardinal: ops... sorry!! =p –  Yuki Jun 8 '12 at 2:12

Let $\lambda_j, j=1 \ldots k$ be the distinct eigenvalues of $U$ (which must be numbers of absolute value $1$). For each $\lambda_j$ let $\mu_j$ be a square root of $\lambda_j$. These also have absolute value $1$. There is a polynomial $p(z)$ such that $p(\lambda_j) = \mu_j$ for each $j$. Let $S = p(U)$.

1) $S^2 = p(U)^2 = U$: in fact $p(z)^2 - z$ is divisible by $\prod_j (z - \lambda_j)$, which is the minimal polynomial of $U$.

2) Since $U$ is normal, the algebra generated by $U$ and $U^*$ is commutative, and in particular $S$ is normal. Since $S$ is normal and its eigenvalues, which are the $\mu_j$, have absolute value $1$, $S$ is unitary.

3) Any nonnegative integer power of a symmetric matrix is symmetric; $S$ is symmetric because it is a linear combination of the symmetric matrices $U^j$.

4) Every matrix that commutes with $U$ commutes with each $U^j$ and therefore with $S$.

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Thank you. How do I find (or prove existence of) such polynomial p(z)? –  user1111261 Jun 23 '12 at 7:06
@user1111261: Lagrange interpolation. –  Robert Israel Jun 24 '12 at 6:44

First, solve for the case of diagonal matrices, which shouldn't be too hard. Then, prove $U$ is diagonalisable, and see if you can use that result to reduce to the previous case.

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Hmm, I had assumed that by $U^*U$ you meant $U$ times $U$, but I guess you mean conjugation... I think this answer is still okay, but I'm not sure. –  Ben Millwood Jun 4 '12 at 17:36
Why is $U$ necessarily diagonalizable? –  copper.hat Jun 4 '12 at 19:40
Every unitary matrix is diagonalizable. en.wikipedia.org/wiki/Unitary_matrix –  user1111261 Jun 4 '12 at 20:03
Ah, good, I thought so :P –  Ben Millwood Jun 4 '12 at 20:14