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

This question is age-old, but it lacks a satisfying answer due to the OP’s omission of some key background information, though Robert gives some clarification along this line in the comments under his answer. So this note is meant to highlight what is touched upon there (just for part (a)): As is stated in the duplicate question On the linear operator $T(M)=AM-MA$, to give a proof, the reader is expected to utilise the continuity of a linear operator $$T$$.

The relevant fact is stated in proposition 5.2.2 in Artin's book:

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

(a)  There is a sequence of matrices $$A_k$$ that converges to $$A$$, and such that for all $$k$$ the characteristic polynomial $$p_k(t)$$ of $$A_k$$ has distinct roots.

[etc.]

Hence, the complex matrix $$A$$ in the question clearly gives rise to such a sequence, the linear operator $$T_k$$ defined by $$T_k(M)=A_kM-MA_k$$ has a rank $$\le n^2-n$$, since $$e_{ii}$$ for $$1 \le i \le n$$ are in its kernel :$$A_k^{-1}T_k(e_{ii})=e_{ii}-A_k^{-1}e_{ii}A_k=0$$.

Now back to our $$T(M)=AM-MA$$, the operator depends continuously on $$A$$, therefore by continuity, $$\forall k( \rm rank \it(T_k) \le n^2-n) \implies \rm rank \it(T) \le n^2-n$$.

• @SungjinKim I don’t understand, seeing we are expected to find the maximum rank, it doesn’t matter if it turns out to be less than that. Commented Jan 5 at 0:55
• Yes, I realized it later, I removed my comment. Commented Jan 5 at 0:56
• @SungjinKim It’s actually okay to keep it, for people often get through quite narrowly without noticing the danger, we need constant poking from others to really attain clarity and soundness :) Commented Jan 5 at 0:59
• My comment was: The continuity argument does not work for rank equality, $1\times 1$ matrix $A_k=(1/k)$ already gives the example. rank($A_k$)$=1$ but rank($\lim_k A_k$)$=$rank$(0)=0$. Commented Jan 5 at 1:48
• @SungjinKim Thank you for bringing it back, my first observation is that $T$ is not any arbitrary operator, so I don’t think we can pick a matrix that special. In particular, is there a matrix $B$ s.t. $BM-MB=AM$ with $A$ being the matrix you give? Commented Jan 5 at 2:53