I gave a proof but it seems rather redundant. What am I doing wrong?
Given $A\in F^{n\times n}$ we define an operator $T_A:F^n \rightarrow F^n$ as such: $$T_A(v) = A \cdot v$$
Prove that $A$ is diagonalizable, iff $T_A$ is diagonizable.
My proof:
- If $T_A$ is diagonizable, then there exists a basis $C$ s.t. $[T_A]_C$ is a diagonal matrix. Since $A$ is the representative matrix of $T_A$ in the standard basis, we can infer that $$A=PCP^{-1}$$ Since A can be represented as $A=PCP^{-1}$, by definition, it is diagonizable.
- If $A$ is diagonizable, and $A$ is the representative matrix of $T_A$ in the standard basis, by definition $T_A$ is diagonizable as well.