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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 $S$
Gerry Myerson
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user1111261
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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: http://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Babylonian_method – Giuseppe Negro Jun 04 '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 06 '12 at 02:21
  • I would be very grateful if you did, @cardinal – user1111261 Jun 06 '12 at 16:03
  • @Yuki: Complex symmetric matrices are not necessarily diagonalizable. I have some patching to do. :) – cardinal Jun 08 '12 at 02:03
  • @cardinal: ops... sorry!! =p – Yuki Jun 08 '12 at 02:12
  • @cardinal I Thinkl complex symmetric matrices (hermitian) are diagonalizable because of the spectral theorem.. – asdf Jun 09 '20 at 17:05

3 Answers3

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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 $S$.

Robert Israel
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    Thank you. How do I find (or prove existence of) such polynomial p(z)? – user1111261 Jun 23 '12 at 07:06
  • @user1111261: Lagrange interpolation. – Robert Israel Jun 24 '12 at 06:44
  • I have a followup-questions regarding this here, if someone can help me. – HerpDerpington Mar 21 '24 at 20:43
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    The polynomial doesn't necessarily exist for a given choice of square roots of the eigenvalues, since it is possible that $\lambda_i=\lambda_j$ while the choices of square roots have $\mu_i\ne \mu_j$. As a consequence, the claimed results are true for a particular square root matrix $S$, but not necessarily all matrices $S$ with $S^2=U$. – Joshua Tilley Mar 21 '24 at 22:13
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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.

Ben Millwood
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Another simple proof based on functional calculus: every unitary is of the form $U=\sum_{j=1}^kc_kE_j$ where $E_j$ are orthogonal projections fulfilling $\sum_jE_j=I$ and $E_jE_{j'}=\delta_{jj'}E_j$, while $c_j\in\mathbb C$, $|c_j|=1$. Its adjoint is $U^*=\sum_j\bar c_jE_j$; it fulfils $U=U^*$ if and only if $c_j=\bar c_j$, namely $c_j=\pm 1$. Define $s_j=\sqrt{c_j}\in\{1,i\}$ and $S=\sum_js_jE_j$. Since each $|s_j|=1$, $S$ is unitary. $S^2=\sum_{jj'}s_js_{j'}E_jE_{j'}=\sum_{j}s_j^2E_j=\sum_jc_jE_j=U$. Since every matrix commuting with $U$ is of the form $A=\sum_ja_jE_j$, $SA=\sum_{jj'}a_jc_{j'}E_{j}E_{j'}=\sum_ja_jc_jE_j$ and analogously $AS=\sum_{j}c_ja_jE_j$.

Gherardo
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