$\DeclareMathOperator\GL{GL}$Let $G$ be a subgroup of $\GL_n(\mathbb{C})$ such that for every $g \in G$ there exists $c \in \GL_n(\mathbb{C})$ for which $cgc^{-1}$ is unitary (or, which is the same, $g$ is diagonalizable with unimodular eigenvalues). Does there exist $c \in \GL_n(\mathbb{C})$ such that $cGc^{-1}$ consists of unitary matrices?

A similar question is discussed in Venkataramana's answer to Groups of matrices in which all elements have all eigenvalues equal in modulus, and this way I can obtain that $G$ is isomorphic to a subgroup of the unitary group.

Let $\mathbb{R}^n = V$, and $0 = V_0 \subset V_1 \subset V_2 \subset \dotsb \subset V_k = V$ be subspaces such that each $V_i$ is invariant under $G$ and the induced action of $G$ on $V_{i+1}/V_i$ is irreducible. Then I get a homomorphism: $G \to \GL(V_1 \oplus (V_2/V_1) \oplus \dotsb \oplus (V_{k}/V_{k-1}))$, and (see the link above) I can assume that its image consists of unitary matrices. Its kernel is trivial, since this homomorphism preserves eigenvalues and the kernel consists of diagonalizable matrices with all eigenvalues equal to $1$.

Now I see two ways:

  1. This homomorphism (denoted as $S$ and seen as a bijection: $G \to \operatorname{Im}S$) is continuous. If $S^{-1}$ is continuous, I think I'm able to prove that $G$ is bounded. But I'm not sure that it's continuous.

  2. $\operatorname{tr}(ab)$ was a non-degenerate form on $\operatorname{Mat}_{n \times n}(\mathbb{C})$. Will it be non-degenerate on a subspace generated by $G$?

  3. Something else….

  • $\begingroup$ If all operators in $G$ are uniformly bounded, then $G$ preserves the norm $\|x\|:=\sup_{g\in G} |gx|$ (where $|x|$ is some fixed norm). Thus it preserves the John ellipsoid of the unit ball of $\|\cdot\|$, that yields affirmative answer. $\endgroup$ May 25, 2023 at 20:12
  • $\begingroup$ Fedor Petrov, thank you for noting this! The hard part is to prove that the operators are bounded. $\endgroup$ May 25, 2023 at 20:18
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    $\begingroup$ The conclusion is negative as [Dave Benson's answer] shows. However, I think it can be shown that one can block-triangulate so that diagonal blocks are unitary. In particular, the conclusion is positive in the irreducible case. $\endgroup$
    – YCor
    May 25, 2023 at 21:25

3 Answers 3


An example in $GL(3, {\mathbb C})$ was first given by Bass (answering a question by Kaplansky) in Example 1.10 of

Bass, Hyman, Groups of integral representation type, Pac. J. Math. 86, 15-51 (1980). ZBL0444.20006.

In section 1 of his paper Bass also discusses the general structure of subgroups of $GL(n, {\mathbb C})$ where every element is unitarizable.

Here is the example. Start with a free subgroup $F=\langle s, t\rangle$ of rank 2 in $SU(2)$ acting linearly on ${\mathbb C}^2$ with generators acting as matrices $A, B$. Now, deform $F$ to a free group of affine transformations by $$ \rho(s)=A, \rho(t)x= Bx+ v, $$ where $v$ is any nonzero vector in ${\mathbb C}^2$. Then each element of $G=\rho(F_2)$ is affine-conjugate to its linear part, an element of $SU(2)$. At the same time, $G$ has no fixed points in the complex-affine space ${\mathbb C}^2$. It follows that $G$ is unbounded. Lastly, use the fact that the group of complex-affine transformations of ${\mathbb C}^2$ embeds in $GL(3, {\mathbb C})$ sending an affine transformation $$ x\mapsto Mx + v $$ to the matrix $$ \left[\begin{array}{cc} M&v\\ 0&1 \end{array}\right]. $$ Under this map unbounded subsets map to unbounded subsets. This gives an example of a non-unitarizable subgroup of $SL(3, {\mathbb C})$ such that each individual element is unitarizable.


Take the group of matrices of the form $\left(\begin{smallmatrix}A&B\\0&A^{-1}\end{smallmatrix}\right)$ with $A\in U(2)$. Inside this, choose two generators for a free group as the values of $A$, in such a way that no two words or their inverses give the same eigenvalues, and with two independent scalar matrices as values of $B$, and see what they generate; in the group generated by these two matrices, all elements are diagonalisable since they don't have a repeated eigenvalue (apart from the identity), but this group is not bounded so it can't be conjugated into a unitary group.

Edit (26 May 2023) in response to comments:

Presumably the example of Bass (I don't seem to be able to find it in his writings) is similar, but using the group $G$ of matrices of the form $\left(\begin{smallmatrix}A&b\\0&1\end{smallmatrix}\right)$ with $A\in U(2)$ and $b\in\mathbb{C}^2$. This is similar but easier, so let me discuss this. Choose a pair of elements of $U(2)$ close to the identity, generating a dense free subgroup consisting of semisimple elements. Since the Lie algebra of $G$ is generated by two elements, we may attach vectors to these two generators so that the subgroup of $G$ generated is dense, hence unbounded.

  • $\begingroup$ Sorry about the multiple edits, it took me a while to get the example working properly. I think it should be okay now. $\endgroup$ May 25, 2023 at 20:58
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    $\begingroup$ You can even lower the dimension to 3 (the example is due to Bass). $\endgroup$ May 25, 2023 at 20:58
  • $\begingroup$ I was originally trying dimension 2, and - surprise, surprise - it didn't work properly. $\endgroup$ May 25, 2023 at 21:11
  • $\begingroup$ What is the 3-dimensional example? $\endgroup$ May 25, 2023 at 21:32
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    $\begingroup$ Ah! I didn't see @MoisheKohan's response above. Yes, this is essentially the example of Bass. $\endgroup$ May 26, 2023 at 10:19

It does not seem to have been mentioned in your question, or in the answer and comments that a periodic subgroup $G$ of ${\rm GL}(n,\mathbb{C})$ satisfies your condition. There were several theorems of Jordan, Burnside and Schur which culminated in the theorem of Schur that a periodic subgroup of ${\rm GL}(n,\mathbb{C})$ is similar to a group of unitary matrices. See Section 36 of chapter V of Curtis and Reiner (1962).


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