$A$ is some fixed matrix. Let $U(B)=AB-BA$. If $A$ is diagonalizable then so is $U$?

Question

This is from Hoffman and Kunze 6.4.13. I am studying for an exam and trying to solve some problems in Hoffman and Kunze.

Here is the question. Let $V$ be the space of $n\times n$ matrices over a field $F$. Let $A$ be a fixed matrix in $V$. Let $T$ and $U$ be linear operators on $V$ defined by

$T(B)=AB$ and $U(B)=AB-BA$.

a) (True or False) If $A$ is diagonalizable then $T$ is diagonalizable. This is true. I can show that both A and T have the same minimal polynomial. So if $A$ is diagonalizable then the minimal polynomial of $T$ should be a product of distinct linear factors. Proving that $T$ is diagonalizable. but its the next question I am having trouble with.

b)(True or false) If $A$ is diagonalizable then $U$ is diagonlizable. I am thinking this is false. But I can't think of of a counter example. May be I am wrong.

Can anybody help?. Thanks for all your help.

score 22 · Accepted Answer · edited Apr 18 '20 at 21:45

22

(a) True. Indeed $p(T)B = p(A)B$ for every polynomial in $F[X]$ whence $p(T) = 0$ if and only if $p(A)=0$. By the minimal polynomial characterization of diagonalizability in finite dimension (splits with simple roots), it follows that $T$ is diagonalizable if and only $A$ is diagonalizable.

(b) True. If $A=P^{-1}DP$ is diagonalizable, then $U$ is similar to $V(B)=DB-BD$ with $D=\mbox{diag}(d_1,\ldots,d_n)$ diagonal. Precisely, $U=S\circ V\circ S^{-1}$ with $S(B)=P^{-1}BP$.

Then denoting $E_{ij}$ the canonical basis of $M_n(F)$, we find $$V(E_{ij})=DE_{ij}-E_{ij}D=(d_i-d_j)E_{ij}.$$ Whence $V$ is diagonal in the canonical basis, and $U$ is therefore diagonalizable.

edited Apr 18 '20 at 21:45

user0

3,567

answered Jun 16 '13 at 03:41

Julien

45,674

Thanks you for your answer. This is helpful. – Cousin Jun 16 '13 at 03:47
While taking $S^{-1}$ one should first make sure that inverse of $B$ exists. Is it true ? – Dhrubajyoti Sarkar Nov 04 '17 at 13:19
@jaggu What about converse i.e. diagonalizablity of $T$ implies that of $A?$ – neelkanth Mar 29 '18 at 10:40
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@jaggu there is no need for $B$ to be invertible. $S^{-1}(B) = PBP^{-1}$ as $SS^{-1}(B) = B$. – Majid May 22 '18 at 21:08
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To help readers avoid confusion, I comment that $V$ is doing double duty: In the question, $V$ is a vector space. In the answer, $V$ is a linear operator. – user0 Apr 18 '20 at 18:56

score 4 · Answer 2 · answered Jun 16 '13 at 08:28

Diagonalise $A$ as $PDP^{-1}$. Then the matrix of $U$ w.r.t. the canonical basis of $M_n(F)$ is $I\otimes A - A^T\otimes I$. Therefore $$ (P^T \otimes P^{-1}) (I\otimes A - A^T\otimes I) (P^{-T} \otimes P) = I\otimes D - D\otimes I $$ is diagonal, i.e. $U$ is diagonalisable.

This answer, of course, requires knowledge in Kronecker product. As your question comes from a Hoffman and Kunze exercise, you are not expected to use Kronecker product, but to solve the question in a linear algebraic way. However, I think this exercise is a perfect example that sometimes a matrix theoretic solution can be much neater than a linear algebraic one.

$A$ is some fixed matrix. Let $U(B)=AB-BA$. If $A$ is diagonalizable then so is $U$?

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