After solving a similar problem in Hoffman and Kunze, I'm trying this one for a long time, but I didn't get success. I tried use the fact that the eigenvalues of $U$ are $\lambda_i-\lambda_j$ with $1\leq i,j\leq n$ where $A$ is $n$ x $n$ and $\lambda_i$ are the eigenvalues of $A$. Of course, $A$ has an eigenvector because the field is algebraically closed. But, even though knowing these facts I'm troubling in prove the main question about $A$ be diagonalizable. Any suggestions will be appreciated! Thanks, everyone.
Asked
Active
Viewed 160 times
3
Rodrigo de Azevedo
- 23,223
Michel Faleiros
- 143
-
Does this hold for any $B$? – Mostafa Ayaz Sep 08 '19 at 22:03
-
@MostafaAyaz He doesn't mean $U(B)$ is diagonalizable, he means the operator $U$. – N. S. Sep 08 '19 at 22:04
-
Oh. I see. Thanks. – Mostafa Ayaz Sep 08 '19 at 22:05
-
1The claim is true if $A$ is a single Jordan block... This suggests to look at the Jordan cannonical form. – N. S. Sep 08 '19 at 22:16
-
I thought about $U$ has at least $n$ eigenvectors associated with the null eigenvalue of $U$, and thus $AB_k=B_kA$ for at least $n$ distinct eigenvectors of $B_k$ of $U$. Thus using that to yield $n$ distinct eigenvectors to $A$... but it didn't work. – Michel Faleiros Sep 09 '19 at 02:34
1 Answers
1
Hint. The idea here is that you can construct a generalised eigenspace of $U$ from a generalised eigenspace of $A$. Suppose $A$ is not diagonalisable. Then for some eigenvalue $\lambda$ of $A$, we have $Ax=\lambda x$ and $Ay=\lambda y+x$ for some eigenvector $x$ and some generalised eigenvector $y$. Now, let $z$ be a left eigenvector of $A$. Define $B=xz^T$ and $C=yz^T$. What are $U(C)$ and $U^2(C)$?
user1551
- 149,263