# The rank and eigenvalues of the operator $T(M) = AM - MA$ on the space of matrices

This problem is from Artin Algebra Second edition, 5.2.3.

Let $A$ be a $n\times n$ complex matrix.

(a) Consider the linear operator $T$ defined on the space $\mathbb{C}^{n\times n}$ of all complex $n\times n$ matrices by the rule $T(M) = AM - MA$. Prove that the rank of this operator is at most $n^2-n$

(b) Determine the eigenvalues of $T$ in terms of the eigenvalues $\lambda_1,\cdots,\lambda_n$ of $A$.

For the part (a), I tried to use Dimension Formula. But, I don't know how to find $\dim(\ker(T))$ is greater than equal to $n$.

For the part (b), I really don't know...

Can someone help me?

• Note that all powers of $A$ are in the kernel. – Gerry Myerson Oct 12 '14 at 6:49
• But they need not be independent linearly. – Abishanka Saha Oct 12 '14 at 9:23
• True, but you have to start somewhere. – Gerry Myerson Oct 12 '14 at 12:04
• It suffices to prove that any Jordan form commutes with at least $n$ l.i. matrices.... – N. S. Oct 13 '14 at 17:37
• I'm sorry, what is "n l.i. matrices"? – baek Oct 14 '14 at 0:27

Hint: if $A$ is diagonal, things are rather simple. Diagonalizable matrices are dense...

• I'm not sure what you are saying...Can you be more specific about your hint? – baek Oct 13 '14 at 5:15
• The rank and the eigenvalues are invariant under similarity, so if $A$ is diagonalizable we may assume $A$ is diagonal. If $A$ is diagonal, the matrix $E_{ij}$ with the $(i,j)$ entry $1$ and the others $0$ is an eigenvector of $T$ for eigenvalue $A_{ii} - A_{jj}$ – Robert Israel Oct 13 '14 at 6:55
• If a sequence of matrices $A_j \to A$, the corresponding operators $T_j \to T$; the rank of $T$ is at most the lim inf of the ranks of the $T_j$, and any eigenvalue of $T$ is a limit point of eigenvalues of $T_j$. – Robert Israel Oct 13 '14 at 7:01
• If A is diagonal matrix, like you said, we can get eigenvalues for T. But, I'm still stuck with general cases. By the way, what do you mean by "the rank of T is at most the lim inf of the ranks of the Tj, and any eigenvalue of T is a limit point of eigenvalues of Tj"? – baek Oct 14 '14 at 1:35
• $\text{Rank}(T) \le \liminf_{j \to \infty} \text{Rank}(T_j)$. In this case (ranks being discrete) it means that for infinitely many $j$, $\text{Rank}(T) \le \text{Rank}(T_j)$. Or in other words, if $\text{Rank}(T_j) \le m$ for all sufficiently large $j$, then $\text{Rank}(T) \le m$. – Robert Israel Oct 14 '14 at 1:43

If $A$ is diagonalizable then we can let $$A = {\rm diag}\ (\lambda_1,\cdots , \lambda_n)$$

If $e_{ij}$ is a matric whose only nonzero entry is $(i,j)$-entry and its value is $1$, then $$[e_{aa},e_{ia}]=-e_{ia},\ [ e_{ii},e_{ia}]=e_{ia}\ (i\neq a)$$

That is $T$ is diagonalizable and $$T(e_{ia})=(-\lambda_a+\lambda_i)e_{ia}$$

That is $\{ e_{ii}\}$ is in kernel space.