Self-adjoint operator is diagonalisable

Question

I am revising adjoints for a linear algebra exam and am confused as to how to prove this.

Suppose that $T: V \rightarrow V$ has the property that $T^*=aT$ for some complex a. How then do you prove from this that $T$ is diagonalisable?

I don't know if you have to find the matrix representation for the matrices, and prove from there, or if there is some property from finding that the eigenvalues are real?

Any help would be appreciated. Thanks.

Answer 1

The operator is normal, then diagonalizable.