1. ## Inner Product

I'd be grateful for any assistance.

V is a finite dimensional vector space over F (where F is R or C).

Let V be a real vector space and let T be a linear operator on V. Prove that if V has a basis consisting of eigenvectors for T, then there is an inner product on V for which T is self-adjoint.

2. Originally Posted by arg
I'd be grateful for any assistance.

V is a finite dimensional vector space over F (where F is R or C).

Let V be a real vector space and let T be a linear operator on V. Prove that if V has a basis consisting of eigenvectors for T, then there is an inner product on V for which T is self-adjoint.
Let $\{e_1,\ldots,e_n\}$ be a basis consisting of eigenvectors for T, and define an inner product by $\langle x,y\rangle = \sum x_iy_i$, where $x = \sum x_ie_i$, $y = \sum y_ie_i$. Note that this will make T selfadjoint (as you can easily check) provided that V be a real vector space. But the result is false over a complex vector space.

3. Originally Posted by Opalg
Let $\{e_1,\ldots,e_n\}$ be a basis consisting of eigenvectors for T, and define an inner product by $\langle x,y\rangle = \sum x_iy_i$, where $x = \sum x_ie_i$, $y = \sum y_ie_i$. Note that this will make T selfadjoint (as you can easily check) provided that V be a real vector space. But the result is false over a complex vector space.

Thank you. However, I can't see how to show that T is self-adjoint.

4. Originally Posted by arg
Thank you. However, I can't see how to show that T is self-adjoint.
If $\lambda_i$ is the eigenvalue for the eigenvector $e_i$, then $\langle Te_i,e_i\rangle = \langle e_i,Te_i\rangle = \lambda$. If i≠j then $\langle Te_i,e_j\rangle = \langle e_,Te_j\rangle = 0$. It follows by linearity that $\langle Tx,y\rangle = \langle x,Ty\rangle$ for all x and y.

5. Originally Posted by Opalg
If $\lambda_i$ is the eigenvalue for the eigenvector $e_i$, then $\langle Te_i,e_i\rangle = \langle e_i,Te_i\rangle = \lambda$. If i≠j then $\langle Te_i,e_j\rangle = \langle e_,Te_j\rangle = 0$. It follows by linearity that $\langle Tx,y\rangle = \langle x,Ty\rangle$ for all x and y.

How do we know that $\langle Te_i,e_j\rangle = \langle e_,Te_j\rangle = 0$?

6. Originally Posted by arg
How do we know that $\langle Te_i,e_j\rangle = \langle e_,Te_j\rangle = 0$?
Because $Te_i=\lambda e_i$, and $\langle e_i,e_j\rangle=0$ by the way in which the inner product was defined.

7. Thanks!