# Inner Product

• Dec 12th 2008, 12:05 PM
arg
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.
• Dec 13th 2008, 03:48 AM
Opalg
Quote:

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 $\displaystyle \{e_1,\ldots,e_n\}$ be a basis consisting of eigenvectors for T, and define an inner product by $\displaystyle \langle x,y\rangle = \sum x_iy_i$, where $\displaystyle x = \sum x_ie_i$, $\displaystyle y = \sum y_ie_i$. 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.
• Dec 13th 2008, 10:02 AM
arg
Quote:

Originally Posted by Opalg
Let $\displaystyle \{e_1,\ldots,e_n\}$ be a basis consisting of eigenvectors for T, and define an inner product by $\displaystyle \langle x,y\rangle = \sum x_iy_i$, where $\displaystyle x = \sum x_ie_i$, $\displaystyle y = \sum y_ie_i$. 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.
• Dec 13th 2008, 10:26 AM
Opalg
Quote:

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

If $\displaystyle \lambda_i$ is the eigenvalue for the eigenvector $\displaystyle e_i$, then $\displaystyle \langle Te_i,e_i\rangle = \langle e_i,Te_i\rangle = \lambda$. If i≠j then $\displaystyle \langle Te_i,e_j\rangle = \langle e_,Te_j\rangle = 0$. It follows by linearity that $\displaystyle \langle Tx,y\rangle = \langle x,Ty\rangle$ for all x and y.
• Dec 13th 2008, 03:25 PM
arg
Quote:

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

How do we know that $\displaystyle \langle Te_i,e_j\rangle = \langle e_,Te_j\rangle = 0$?
• Dec 14th 2008, 12:48 AM
Opalg
Quote:

Originally Posted by arg
How do we know that $\displaystyle \langle Te_i,e_j\rangle = \langle e_,Te_j\rangle = 0$?

Because $\displaystyle Te_i=\lambda e_i$, and $\displaystyle \langle e_i,e_j\rangle=0$ by the way in which the inner product was defined.
• Dec 14th 2008, 08:35 AM
arg
Thanks!