# Group of matrices in which every matrix is similar to unitary

$$\DeclareMathOperator\GL{GL}$$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….

• 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. May 25 at 20:12
• Fedor Petrov, thank you for noting this! The hard part is to prove that the operators are bounded. May 25 at 20:18
• 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.
– YCor
May 25 at 21:25

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.

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