Eigenvalues and rank of $T(X)=AX$

Question

$T:M_n(\mathbb{C})\rightarrow M_n(\mathbb{C})$ defined by $T(X)=AX$ where $A$ is a fixed $n\times n$ matrix. I wish to find the characteristic polynomial and the rank of $T$. I looked at where $T$ sends the basis elements $E_{lm}$. I found that $T(E_{lm})=\sum\limits_k a_{kl}E_{km}$. From here I calculated the trace to be $n\operatorname{trace}(A)$. Finding the characteristic polynomial seemed to be difficult. I calculated for $n=2$. Letting $$A= \bigg(\begin{matrix} a & b \\ c & d \end{matrix}\bigg)$$ I got $\chi_T(x)=(x-a)(x-d)\chi_A(x)$ where $\chi$ denotes the characteristic polynomial. However, I don't know how this generalizes to higher dimensions. The diagonal entries and characteristic roots of $A$ only account for $2n$ eigenvalues and not $n^2$. Also what about the rank of $T$? Does it only depend on the rank of $A$? By the looks of it $\operatorname{rank}T$ could also depend on how many diagonal entries of $A$ are $0$. Any help and/or reference would be appreciated.

Edit: I made an error; $\chi_T(x)$ is actually $\chi_A(x)^2$

Does this answer your question? A linear map $T$ in $M_n(K)$ — Dietrich Burde, Dec 07 '19 at 15:18
@DietrichBurde Sorry, but no. They assume $A$ to be invertible, which is a very special case and gives the rank immediately. Also it says nothing about the eigenvalues. — Not Euler, Dec 07 '19 at 15:45
It would improve your Question to format it in a way that highlights your problems(s). You want some information about the linear map induced by multiplication by $A$, but it is visually difficult to find the specific information you are asking for. — hardmath, Dec 16 '19 at 17:50

score 1 · Accepted Answer · answered Dec 07 '19 at 15:51

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I'll do the $2\times 2$ case. With respect to the basis $$E_1=\pmatrix{1&0\\0&0},\quad E_2=\pmatrix{0&0\\1&0},\quad E_3=\pmatrix{0&1\\0&0},\quad E_4=\pmatrix{0&0\\0&1}$$ the transformation $T$ has matrix $$A'=\pmatrix{a&b&0&0\\c&d&0&0\\0&0&a&b\\0&0&c&d}.$$ It's apparent that the characteristic polynomial of $A'$ is the square of that of $A$.

In the $n$-by-$n$ case we get $n$-th powers by a similar argument.

answered Dec 07 '19 at 15:51

Angina Seng

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One could mention that "under the hood", we work with Kronecker product(s) $\otimes$, here $vec(AX)=(I_2 \otimes A)vec(X)$ – Jean Marie Dec 08 '19 at 07:51

Eigenvalues and rank of $T(X)=AX$

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