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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?

PinkyWay
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baek
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4 Answers4

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

Robert Israel
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  • I'm not sure what you are saying...Can you be more specific about your hint? – baek Oct 13 '14 at 05:15
  • 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 06:55
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    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 07:01
  • 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 01:35
  • $\text{Rank}(T) \le \liminf_{j \to \infty} \text{Rank}(T_j)$. 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 01:43
  • A way to see this is that if $\text{Rank}(T) \ge m+1$, there would be some $m+1 \times m+1$ submatrix $S$ of $T$ that is invertible. This would be the limit of the corresponding submatrices $S_j$ of $T_j$. But invertible matrices form an open set. – Robert Israel Oct 14 '14 at 01:48
  • Any eigenvalue $\lambda$ of $T$ is the limit of a sequence $\lambda_j$, where $\lambda_j$ is an eigenvalue of $T_j$. This follows from Hurwitz's theorem of complex analysis, or from the argument principle. – Robert Israel Oct 14 '14 at 01:57
  • The rank can change actually, since we are considering the space of $n\times n$ matrices...But it's safe to claim that the similar transformation preserves the linear independency. – Bach Jan 10 '20 at 21:09
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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.

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

C_Arietta_C
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  • @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. – C_Arietta_C Jan 05 '24 at 00:55
  • Yes, I realized it later, I removed my comment. – Sungjin Kim Jan 05 '24 at 00:56
  • @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 :) – C_Arietta_C Jan 05 '24 at 00:59
  • 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$(0)=0$. – Sungjin Kim Jan 05 '24 at 01:48
  • @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? – C_Arietta_C Jan 05 '24 at 02:53
  • @SungjinKim But, back to the question, you did make me realize that concerning continuity, it’s better to reason in terms of the eigenvalues, now obviously $T_k$ has eigenvalues $\lambda_{k,1}$ through $\lambda_{k,n}=0$, by continuity of eigenvalues (also included in proposition 5.2.2, viz. (c)), $T$’s eigenvalues $\lambda_1$ through $\lambda_n$, being limits of constant sequences, cannot be other than that constant, i.e. $0$. – C_Arietta_C Jan 05 '24 at 02:53
  • Your argument is fine, no worries. My comment was for rank equality. Yours is rank inequality $\le$ so there is no problem with yours. – Sungjin Kim Jan 05 '24 at 05:40
  • @SungjinKim I’d rather say that your comment really hit home, I was quite unclear as to what exactly is meant by continuity in the case of rank, and thanks to your doubt came to realize that in order to say anything about the rank in the light of continuity, it is the kernel and hence the zero eigenvalues that need to be studied, which also means equality is out of the question. – C_Arietta_C Jan 05 '24 at 05:59
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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.

PinkyWay
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HK Lee
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