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Let $n\ge 2$ and $A\in GL(n,\mathbb R)$ be a matrix and define a linear transformation on $M(n,\mathbb R)$ as follows:

$$f_A: M(n,\mathbb R) \to M(n,\mathbb R), B \to ABA^{-1}. $$

Suppose $f_A$ is a unipotent linear transformation on $M(n,\mathbb R) \cong \mathbb R^{n^2}$, namely the only eigenvalue of $f_A$ is $1$. Do we necessarily have that the matrix $A$'s eigenvalues are the same? (scalar multiple of a unipotent)

Conversely, if $A$ is a unipotent matrix, do we necessarily have that $f_A$ is a unipotent linear transformation?

Update: Many thanks to the comments below about the issue of unipotency, corrected a little.

No One
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  • Take $A$ to be a scalar multiple of the identity. – Qiaochu Yuan Aug 10 '20 at 07:42
  • We can avoid that somewhat pathological example by requiring merely that all eigenvalues are the same, or that up to scalars it's a unipotent transformation. (That also eliminates the failure of the case $n=1$.) – David A. Craven Aug 10 '20 at 08:40

1 Answers1

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I've started to write an answer to this twice now, so I'm going to make sure I finish this time.

The action of $A\in \mathrm{GL}_n(k)$ on $M_n(k)$ is given by a Kronecker product. The precise product is the Kronecker product of $A$ with its dual $A^*$, which can be thought of as $(A^{-1})^t$, where $t$ denotes the transpose. We will see that the precise form of the Kronecker product isn't so important.

The eigenvalues of the Kronecker product are simply all products of the eigenvalues of the matrices. If the eigenvalues of the product are all $1$, then the eigenvalues of each factor are all the same. You see that you don't need to know exactly what the other factor was, just that $A$ was a factor, to determine the one direction of your question.

Conversely, if $A$ is unipotent, then so is $A^*$, and so their Kronecker product is as well. Notice that even if we replace $A$ by its product with some scalar, the Kronecker product is the same, as it should be.

(There is a more conceptual answer involving Lie theory, but I guess you didn't want that since you didn't couch the question in those terms.)

David A. Craven
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  • The action of A is by conjugation.... how is this related to the Kronecker product you mentioned? – No One Aug 10 '20 at 23:23