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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?

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

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  • $\begingroup$ I'm not sure what you are saying...Can you be more specific about your hint? $\endgroup$ – baek Oct 13 '14 at 5:15
  • $\begingroup$ 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 $A_{ii} - A_{jj}$ $\endgroup$ – Robert Israel Oct 13 '14 at 6:55
  • $\begingroup$ 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 $T_j$. $\endgroup$ – Robert Israel Oct 13 '14 at 7:01
  • $\begingroup$ 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"? $\endgroup$ – baek Oct 14 '14 at 1:35
  • $\begingroup$ $\text{Rank}(T) \le \liminf_{j \to \infty} \text{Rank}(T_j)$. 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 $\text{Rank}(T) \le m$. $\endgroup$ – Robert Israel Oct 14 '14 at 1:43
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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 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.

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