# Let $V$ an inner product space and $T\in\mathcal L(V)$ normal. Show that the minimal polynomial of $T$ have no repeated zeros

This is the exercise 13 in page 268 of Linear algebra done right, third edition of Axler.

Let $V$ an inner product space and $T\in\mathcal L(V)$ normal. Show that the minimal polynomial of $T$ have no repeated zeros.

Here it is assumed that $V$ is finite-dimensional.

The result, in the context of the book, is immediate if $V$ is a complex finite-dimensional vector space. However if $V$ is a real vector space it is not so clear if it is probable in the context where the exercise is asked.

Every theorem of the chapter, except perhaps some general definitions as the minimal polynomial, are done in the context of $V$ as a complex finite-dimensional vector space.

This question was asked previously here without an answer (the comment on the question seems not correct because an operator in $\Bbb C^n$ doesnt necessarily can be seen as an operator in $\Bbb R^n$, by example $T(x,y)=(ix,x-iy)$ cannot be real.)

Questions:

1. Can someone clarify if the exercise can be solved for real $V$ in the context of the book?

2. Can someone confirm if the statement is true or false for real $V$?

• An operator $\mathbb{R}^n \to \mathbb{R}^n$ can be extended canonically to one $\mathbb{C}^n \to \mathbb{C}^n$. – Chappers Jul 19 '17 at 13:07
• yes @Chappers, but and operator on $\Bbb C^n$ cannot be injected/embedded in $\Bbb R^n$ in the general case. I can prove the statement of the exercise for $\Bbb C^n$ from the theory (and previous exercises) of the chapter but Im not sure if the statement is also probable for operators in $\Bbb R^n$. – Masacroso Jul 19 '17 at 13:17
• The comment on that post is correct: of course, not every operator in $\Bbb C^n$ can be seen as an operator over $\Bbb R^n$; however, every operator over $\Bbb R^n$ can be seen as (i.e. extended to) an operator over $\Bbb C^n$. This extension does not change the minimal polynomial. – Omnomnomnom Jul 19 '17 at 13:24