Distinct eigenvalues implies $A \in \mathbb{R}^{n \times n}$ is diagonalisable

Question

Theorem:

If an $n \times n$ matrix has n distinct eigenvalues then A is diagonalisable.

Proof:

Let $A \in \mathbb{R}^{n \times n}$. Suppose A is not diagonalisable. Then, by definition, for a given $D \in \mathbb{R}^{n \times n}$ there exists no invertible matrix $P \in \mathbb{R}^{n \times n}$ such that $P^{-1}AP = D$.

Any hint(s) to assist me would be helpful.

Thanks in advance.

Answer 1

Hint: I would prefer to prove it directly.

Eigenvectors for distinct eigenvalues are linearly independent. Thus we have a basis for $V$ consisting of eigenvectors. Simply let $P$ be the matrix whose columns are the basis vectors.

Then $P^{-1}AP=D$, where $D$ is diagonal, and the entries on the diagonal are the eigenvalues.