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In this answer the point is made that since a unitary and symmetric matrix $U$ commutes with all $U^j$ (I assume those are the matrix powers $U^2$, $U^3$ etc?) it must also commute with $S=\sqrt{U}$, where $\sqrt{\cdot}$ denotes the matrix squre root. I do not understand this argument. The matrix square root is not necessarily unique. I can see that if $S$ commutes with some matrix, so must $U$, but not necessarily the other way around.

In other words can there (or why can there not) be a matrix $\mathcal{S}$, which does not commute with $U$, but which nevertheless fullfills $\mathcal{S}^2=U$?

HerpDerpington
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    This https://math.stackexchange.com/a/1116701/42969 and this https://math.stackexchange.com/a/2778855/42969 might help. – Martin R Mar 21 '24 at 20:51
  • @MartinR It seems to help, but now I can only rephrase the question: Why am I allowed to choose 1/2 as the exponent for the polynomial here? I don't understand the argument with the limit. Is this a "functional-analysis"/"operator-theory" result? Maybe an answer to this question could also just summarize/state the important theorem(s) here which I would need to understand this. Also: How/why does this positive definite operators? – HerpDerpington Mar 21 '24 at 20:55
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    The missing course-work really comes from complex-analysis. As far as the second paragraph, if $S^2=U$ then $SU= S(S^2)=S^3=(S^2)S=US$ – user8675309 Mar 21 '24 at 21:32
  • @user8675309 That's clever, I wonder why I didn't see that. – HerpDerpington Mar 21 '24 at 21:35

2 Answers2

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In fact the claim is not true, even for unitary, symmetric matrices $U$ and $S$. For example, $$U= \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$$ $$S= \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}$$ are unitary and symmetric, and $S^2=U$, but $U$ commutes with

$$M= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$

while $S$ does not.

Joshua Tilley
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I believe a stronger assumption needed on $U$. Since $U$ is unitary , so by spectral theorem, all its eigenvalue must be normed $1$.

But if there exists some symmetric matrix $S$ which is the "square root" of $U$, this means for any $\textbf{v}\in\textbf{k}^n$, we must have $$\textbf{v}^TU\textbf{v}=\textbf{v}^TSS\textbf{v}=\textbf{v}^TS^TS\textbf{v}=(S\textbf{v})\cdot (S\textbf{v})\ge0$$ which shows $U$ must be at least positive semidefinite (definite). This cannot be guaranteed.

Angae MT
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  • Shouldn't the eigenvalues be arbitrary complex numbers for unitary matrices? – HerpDerpington Mar 21 '24 at 21:40
  • Unitary matrix preserves norm in $\mathbb{C}$, so $|\lambda|=1$ for any eigenvalue $\lambda\in\mathbb{C}$. – Angae MT Mar 21 '24 at 21:43
  • But that still means that the eigenvalues can be of the form $exp(i\phi)$ for some real valued $\phi$. – HerpDerpington Mar 21 '24 at 22:15
  • The matrix $i\operatorname{id}$ is unitary and symmetric, but its eigenvalues are not equal to $\pm 1$. In fact a complex symmetric matrix needn't have real eigenvalues. It is a complex Hermitian matrix or a real symmetric matrix you are thinking of, perhaps? – Joshua Tilley Mar 21 '24 at 22:20
  • Thanks for pointing out the mistake. $U$ should be self-adjoint instead of symmetric to have real eigenvalue. But I think the important part here is $U$ has to be positive definite, so hopefully this mistake does not affect my argument. – Angae MT Mar 21 '24 at 22:38
  • What does it mean for a non-hermitian, complex matrix to be positive definite? Your arguments all make sense to me if we restrict to real matrices. – Joshua Tilley Mar 22 '24 at 00:34