# Symmetric endomorphism

1. Jun 19, 2010

### penguin007

Hi,

We know that if u is a real symetric endomorphism, then u has a real eigenvalue and that u is diagonalizable.
But can we say that u is diagonalizable with only real eigenvalues?

2. Jun 19, 2010

### HallsofIvy

Staff Emeritus
Yes. Since you are talking about eigenvalues, I take it that u is an endomorphism on some vector space (a linear transformation from vector space V to itself). Specifically, if u is symmetric and $\lambda$ is an eigenvalue, then there exist non-zero x such that ux= \lambda x. Further, we can take x to have magnitude 1: <x, x>= 1. Then $\lambda= \lambda<x, x>= <\lambda x, x>= <ux, x>= <x, ux>$ (because u is symmetric). $<x , ux>= \overline{<ux, x>}= \overline{\lambda x, x}= \overline{\lambda}\overline{<x, x>}= \overline{\lambda}$.
(The overline indicates complex conjugation.)

Since $\lambda= \overline{\lambda}$, $\lambda$ is real. That is all eigenvalues are real for a symmetric endomorphism.

3. Jun 19, 2010

### penguin007

I think you forgot an overline in the end, but I can't understand why we have:
<x,ux>=conjugate(<ux,x>) ?

4. Jun 20, 2010

### HallsofIvy

Staff Emeritus
One of the requirements in the definition of "inner product" is that $<u, v>= \overline{<v, u>}$. Of course, if the vector space is over the real numbers, that is the same as "<u, v>= <v, u>" but then your question is trivial.