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I'm not sure how to proceed.

Let us find all possible solutions for the matrix $A$ which commutes with any other matrix $X$. In other words:

$$AX=XA$$

Stating the matrix multiplication explicitly we can conclude that for any $i,j\in\{1,2,3\}$:

$$\sum_{n=1}^3 A_{i,n} X_{n,j} = \sum_{n=1}^3 X_{i,n} A_{n,j}$$

At this point I do not know what to do anymore. How do I proceed?

user157101
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    Try a few upper triangular matrices and see what restrictions they impose on the matrices that commute with them. – user2357112 Jun 14 '14 at 20:42
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    The question title and the question body do not match. In the title, your $X$ must be upper triangular, but in the question body, no such requirement is made. At present, the accepted answer by Berci assumes that $A$ commutes with all square matrices (as opposed to only upper triangular matrices). So, he/she is not answering the question as stated in the title. Anyway, please clarify your question. – user1551 Jun 15 '14 at 06:35
  • Duplicate of the question in the body, but not in the title: https://math.stackexchange.com/q/27808/15416 – Julian Kuelshammer Feb 28 '18 at 12:21

2 Answers2

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The solutions are scalar matrices, that is, $A=\lambda I$ for some $\lambda\in R$. It's easy too see if you notice that the solution is commute with matrices that are upper triangular, have all entries $0$ except one, which is equal $1$.

pointer
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The characteristical polynomial of $A$ is of degree $3$, so it has at least one root over $\Bbb R$, say $\lambda$, which is an eigenvalue of $A$.
Let $v$ be an eigenvector belonging to $\lambda$, i.e. $Av=\lambda v$.

Now, for any $u\ne 0$, let $X$ be (the matrix of) a linear map that swaps $v$ and $u$. By the equations we get $$Au=X^{-1}AXu=X^{-1}Av=X^{-1}(\lambda v)=\lambda u\,,$$ so that $A=\lambda\cdot I$ where $I$ is the identity matrix.

Berci
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  • Wow, I'm pretty surprised by the fact that I initially guessed $\lambda\cdot I$, even though I had not really proved it. – user157101 Jun 14 '14 at 21:21
  • Anyway, it works for general $n\times n$ matrices. If not in $\Bbb R$, $\ A$ will have an eigenvalue in $\Bbb C$ and then the same goes through... – Berci Jun 14 '14 at 21:21