# Where n is an odd positive integer

1. Sep 21, 2009

### squenshl

I'm studying for a test.
In doing one of the old tests and it had a question that I couldn't do.
Let T: Rn $$\rightarrow$$ Rn be an operator on Rn,
where n is an odd positive integer. How do I prove T has at least one eigenvector in Rn

2. Sep 21, 2009

### HallsofIvy

Staff Emeritus
Re: Proof

Well, you can't unless it is specified that T is a linear operator- otherwise "eigenvectors" cannot be defined.

I presume that you know that the "characteristic equation" for an operator on an n dimensional space is an nth order polynomial equation. Since the underlying field is the real numbers, all coefficients of that characteristic equation are real numbers. Now, you know, do you not, that, in that situation, complex roots come in complex conjugate pairs?