8
$\begingroup$

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?

$\endgroup$
6
  • 3
    $\begingroup$ Note that all powers of $A$ are in the kernel. $\endgroup$
    – Gerry Myerson
    Oct 12, 2014 at 6:49
  • 1
    $\begingroup$ But they need not be independent linearly. $\endgroup$
    – QED
    Oct 12, 2014 at 9:23
  • 2
    $\begingroup$ True, but you have to start somewhere. $\endgroup$
    – Gerry Myerson
    Oct 12, 2014 at 12:04
  • 1
    $\begingroup$ It suffices to prove that any Jordan form commutes with at least $n$ l.i. matrices.... $\endgroup$
    – N. S.
    Oct 13, 2014 at 17:37
  • $\begingroup$ I'm sorry, what is "n l.i. matrices"? $\endgroup$
    – baek
    Oct 14, 2014 at 0:27

3 Answers 3

Reset to default
4
$\begingroup$

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

$\endgroup$
8
  • $\begingroup$ I'm not sure what you are saying...Can you be more specific about your hint? $\endgroup$
    – baek
    Oct 13, 2014 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, 2014 at 6:55
  • 1
    $\begingroup$ 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, 2014 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, 2014 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, 2014 at 1:43
0
$\begingroup$

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.

$\endgroup$
1
0
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .