0


This is a Homework question i'm struggling with.

T is a linear map over C.
Prove: if T and T* have the same Eigen-vectors then T is normal.

I've been breaking my back over this one...
A complete solution would be nice but any help would do.
thanks.

Michael

m_dunaevski
  • 3

1 Answers1

1

Note: This is incorrect, but there is a fix which follows roughly the same lines below. I will update the answer at some point. See @trfv's comment and my response below. The fix is to use the Schur decomposition and show (by induction) that the upper triangular part of the decomposition is diagonal (that is, the 'departure from normal' is zero).

Here is an approach:

Let $J = \operatorname{diag} (J_1,...,J_p)$ be the (block diagonal) Jordan form of $T$, and let the sizes of each block be $\gamma_k$.

I claim that $\gamma_k = 1$ for all $k$. For the sake of contradiction, suppose this is not true. Then $v=(0,...,0,e_{1}^T,0,...0)^T$ is an eigenvector (the $e_{1}$ is in the $k$th place corresponding to the block structure) of both $J$ and $J^*$. However, $J_k e_{1} = \lambda e_{1}$ for some $\lambda$, but $J_k^* e_{1} = \lambda e_{1}+e_{2}$, a contradiction.

Hence $J$ is diagonal. Now conclude that $A$ is normal.

copper.hat
  • 172,524
  • you mean that there will be a column of J that is an eigenvector of T but not of T* because of the 1 in the jordan form? – m_dunaevski Jun 16 '15 at 07:30
  • @m_dunaevski: The only eigenvector (modulo scaling) of a Jordan block is $e_1$. When you take the (conjugate) transpose, $e_1$ is not an eigenvector because of the one. This is shown by $J^* e_1= \lambda e_1 + e_2$ which shows that $e_1$ is not an eigenvector of $T^*$. – copper.hat Jun 16 '15 at 15:48
  • @copper.hat - if $J$ is diagonal then $A$ is diagonalizable. But to prove it is normal we must show that $A$ is unitarily diagonalizable. Is it something we can conclude from $J$ being diagonal? If yes - how? – trfv Jun 23 '15 at 12:59
  • @trfv: Good catch, you are correct, there is no guarantee that the basis for a Jordan form is orthonormal. The fix (which I don't have time to add above now) is to use the Schur decomposition and show (by induction) that the upper triangular part of the decomposition is diagonal (that is, the 'departure from normal' is zero). – copper.hat Jun 23 '15 at 16:44