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

Question

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?

Answer 1

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

Answer 2

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.

Answer 3

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.

Answer 4

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, $\forall k( \rm rank \it(T_k) \le n^2-n) \implies \rm rank \it(T) \le n^2-n$.