# Linear Algebra

1. Aug 13, 2009

### LeifEricson

1. The problem statement, all variables and given/known data
let $$T:V \to V$$ be a linear transformation which satisfies $$T^2 = \frac{1}{2} (T + T^*)$$ and is normal. Prove that $$T=T^*$$.

2. Relevant equations

3. The attempt at a solution
I think we should start like this:
Let $$\mathbf{A}=[T]_B$$ be the matrix representation of T in the orthonormal base $$B$$. We look at A as a matrix in the Complex plane. If we prove that for every Eigenvalue, $$\lambda$$, it happens that $$\lambda \in \mathbb{R}$$, then $$T = T^*$$ by a theorem and we finished.
Now, be $$\mathbf{x}$$ an Eigenvector that satisfies $$\mathbf{Ax} = \lambda \mathbf{x}$$, then $$<\mathbf{Ax},\mathbf{x}> = \lambda ||\mathbf{x}||^2$$ and $$\lambda = \frac{<\mathbf{Ax},\mathbf{x}>}{||\mathbf{x}||^2}$$. If only I could show that $$<\mathbf{Ax},\mathbf{x}>$$ is a real number, I solved it.

2. Aug 13, 2009

### Feldoh

Prove that the transformation is hermitian, once you do it's pretty east to prove that eigenvalues of a hermitian transformation are real.

3. Aug 13, 2009

### LeifEricson

It doesn't help me, because I plan to show that the transformation is Hermitian by the following theorem:
"If T is a normal transformation whose Characteristic polynomial can be completely factored into linear factors over $$\mathbb{R}$$, then T is Hermitian".
And then it follows, of course, that T = T*

4. Aug 13, 2009

### LeifEricson

Can someone help me please? I work on it two days. Nothing works. Please. I spent hours.