# 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 an $$n\times n$$ 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 part $$(a)$$, I tried to use Dimension Formula. But, I don't know how to show that $$\dim(\ker(T))$$ is greater than equal to $$n$$.

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

Can someone help me?

• Note that all powers of $A$ are in the kernel. Oct 12, 2014 at 6:49
• But they need not be independent linearly.
– QED
Oct 12, 2014 at 9:23
• True, but you have to start somewhere. Oct 12, 2014 at 12:04
• It suffices to prove that any Jordan form commutes with at least $n$ l.i. matrices.... Oct 13, 2014 at 17:37
• I'm sorry, what is "n l.i. matrices"?
– baek
Oct 14, 2014 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, 2014 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 $A_{ii} - A_{jj}$ Oct 13, 2014 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 $T_j$. Oct 13, 2014 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, 2014 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 $\text{Rank}(T) \le m$. Oct 14, 2014 at 1:43

If $$A$$ is diagonal, $$AM-MA$$ will have all it's diagonal entries $$0$$. So $$\{e_{ij}:i\ne j \}$$will be a spanning set of length $$n^2-n$$ .Since basis is a linearly independent set, its length is less than $$n^2-n$$, so $$\;\dim Im\leqslant n^2-n\;$$ or $$\;\operatorname{rank}\leqslant n^2-n$$.

I don't know how to extend this proof to a general case, but we can use continuity I guess.

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

If $$e_{ij}$$ is a matrix 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.