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Suppose $G_1, \ldots, G_k$ are unitary, Hermitian, and anti-commuting matrices, and assume the same for $F_1, \ldots, F_k$. Suppose these matrices are similar, i.e. there exists $T \in GL_n(\mathbb{C})$ such that $$ G_i = T^{-1} F_i T $$ for all $i \in [k]$. Does there exist $V \in U_n$, where $U_n$ is the group of unitary matrices, such that $$ G_i = V^{-1} F_i V $$ for all $i \in [k]$?

Added later:
Generally, let $\pi, \sigma : G \to U_n(\mathbb{C})$ be two unitary representations of a finite group. Suppose $\pi$, $\sigma$ are equivalent, i.e. there exists $T \in GL_n(\mathbb{C})$ such that $T \pi(g) = \sigma(g) T$ for all $g \in G$. Are $\pi$ and $\sigma$ unitarily equivalent, that is, there exists $T' \in U_n$ such that $T' \pi = \sigma T'$?

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Motivation? Ideas already tried? Context? – Yemon Choi Sep 20 '12 at 19:55
I can not see a solution even when $k = 1$ .... – jsliyuan Sep 20 '12 at 20:00
It seems to be a simple application of polar decomposition. – Mateusz Wasilewski Sep 20 '12 at 20:25
I am not keen on answering anonymous users (anonymous != pseudonymous), but, working over ${\mathbb C}$: If two unitary matrices are similar then they have the same eigenvalues, counting multiplicity. Also, every unitary matrix is unitarily similar to a diagonal matrix. So the answer to this question (and your earlier version) would seem to be "yes". – Yemon Choi Sep 20 '12 at 23:30
Thank you. I see. – jsliyuan Sep 21 '12 at 0:24
up vote 1 down vote accepted

Maybe this question deserves an answer which doesn't use decomposition into irreducible representations. Let $\pi$ and $\sigma$ be equivalent unitary representations. If $T$ intertwines, then so does $T^{\ast}$, because $\pi(g^{-1})=\pi(g)^{\ast}$ (same for $\sigma$). Now we can conclude that $|T|=\sqrt{T^{\ast}T}$ is also an intertwiner. Finally, $T|T|^{-1}$ is the unitary we need.

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Nice. I still am partial to an argument that just uses lin algebra at 1st/2nd course level, though – Yemon Choi Sep 21 '12 at 17:46
It's very clear. Thanks – jsliyuan Sep 21 '12 at 20:36
It seems that there is some problem... Using the notations above, $T \pi=\sigma T$, and $\pi T^* = (\pi^{-1})^* T^* = (T \pi^{-1})^* = (\sigma^{-1} T)^* = T^* \sigma$. Thus, both $T$ and $(T^*)^{−1}$ intertwines. $T|T|^{−1}$ is unitary. However, I can't see why $T|T|^{−1}π=\sigma T|T|^{−1}$, which amounts to $\pi|T|=|T|\pi$. Why this is true? – jsliyuan Sep 21 '12 at 22:06
$|T|$ is a norm limit of polynomials in $T^{\ast}T$ and the latter operator can be easily seen to commute with $\pi$. – Mateusz Wasilewski Sep 21 '12 at 22:28
Could you tell me the definition of "norm limit of polynomials"? I haven't seen it before. Thanks :) – jsliyuan Sep 21 '12 at 22:32

Two similar unitary representations are always unitarily similar if they are irreducible. An isomorphism $V_1 \to V_2$ pulls back the unitary metric on $V_2$ to a unitary metric on $V_1$. To make this unitary, we need to change that metric into the original metric on $V_1$ through a $G$-equivariant map $V_1\to V_1$. But any two unitary metrics on a representation are related by a $G$-equivariant map, so this is always possible.

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If two irreducible unitary representations are similar, then T would be a scalar multiple of the identity by Schur's lemma. What happens if two representations are not irreducible? Did you mean that through a $G$-equivariant map $V_1 \to V_1$, we can make $T$ unitary? And how to find such a $G$-equivariant map? – jsliyuan Sep 21 '12 at 0:28
We can write a unitary metric as a unitary matrix with one row or column for each irreducible subrepresentation, where all entries that correspond to two non-isomorphic representations are zero. Similarly, a G-equivariant map is a matrix where all entries that correspond to two non-isomorphi representations are zero. G-equivariant maps act by simultaneous left multiplication and right conjugate multiplication. The only invariant preserved by this action is the signature of each block. Unitary metrics are always positive-definite, so this is the same for both metrics. – Will Sawin Sep 21 '12 at 1:22
In other words, you take a square root. – Will Sawin Sep 21 '12 at 1:22
Did you mean that the answer is "yes" for irreducible representations, and the answer may be "no" for the others? – jsliyuan Sep 21 '12 at 1:45
No, I mean it's true in general. – Will Sawin Sep 21 '12 at 4:03

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