Find an inner product.

Question

Suppose V is a vector space over $\Bbb C$ (without an inner product), and that $T \in \mathcal L (V )$ is diagonalizable.

Prove that there exists an inner product $\langle ·, · \rangle $ on V such that, with respect to this inner product, T is normal.

My original thought is well we have $T= U D U^{-1} $ so we have $TT^{*}= U D U^{-1} (U D U^{-1})^* = U D U^{-1}(U^{-1})^* D^* U^*$ Where U is an change of basis. if we could prove that $U^* = U^{-1}$ when U is a change of basis matrix then we know that two diagonal matrices commute. then we would have $U D^* D U^*= (U^{-1})^* D^* U^* U D U^{-1} = T^{*}T$ as desired.

Is there some easy way of doing so or a diffrent way to approach the problem? ( we havent technically covered unitary matrices yet)

Answer 1

When you write $T=UDU^{-1}$, you get into unnecessary complications. $T$ is not a matrix, it is a linear transformation. If you want to think of it as a matrix, you get to choose the basis, in which case you can directly choose the right one and the matrix of $T$ is diagonal. $M_T=D$.

All that is left to do is explain why for any basis, there is an inner product that makes that basis orthonormal. This is relatively straightforward.

Answer 2

Despite not covering unitary matrices i still think this is the best solution rather than verifying directly all the properties as you need to know nothing about them in this given argument except what is directly proved to get the result.

Let $T= U D U^{-1} $ Where D is diagonal $U$ and $U^{-1}$ are change of basis to make that happen. Next we notice that $U= [I]_{\alpha}^{\beta} $ we know by theorem that $U^*= ([I]_{\alpha}^{\beta})^*=[I]^{\alpha}_{\beta} $ now let us conider $U U^* $

then we have that $ U U^*= [I]_{\alpha}^{\beta} [I]^{\alpha}_{\beta} =[I]= [I]^{\alpha}_{\beta} [I]_{\alpha}^{\beta} =U^* U $ Inverse's are unique so $ U^* = U^{-1} $

so we have $TT^{*}= U D U^{-1} (U D U^{-1})^* = U D U^{-1}(U^{-1})^* D^* U^*$ Where U is an change of basis and $U^* = U^{-1}$ then we know that two diagonal matrices commute. Hence we have that $U D^* D U^*= (U^{-1})^* D^* U^* U D U^{-1} = T^{*}T$ so $TT^{*} =T^{*}T$ as desired.