If an $n \times n$ matrix $A$ has $n$ eigenvalues and $n$ linearly independent eigenvectors is it diagonlizable?

Question

Is this always true? also if a matrix has less than $n$ eigenvalues but still $n$ linearly independent eigenvectors would it be diagonalizable in this case?

Answer 1

Let the $n$ eigenvalues be $\lambda_1$ through $\lambda_n$ and the eigenvectors be $\vec v_1$ through $\vec v_n$.

Then, we have: $$A\begin{bmatrix}\vec v_1&\vec v_2&\cdots&\vec v_n\end{bmatrix} = \begin{bmatrix}\lambda_1\vec v_1&\lambda_2\vec v_2&\cdots&\lambda_n\vec v_n\end{bmatrix}$$

After, that: $$A\begin{bmatrix}\vec v_1&\vec v_2&\cdots&\vec v_n\end{bmatrix} = \begin{bmatrix}\vec v_1&\vec v_2&\cdots&\vec v_n\end{bmatrix} \begin{bmatrix}\lambda_1&0&\cdots&0\\0&\lambda_2&\cdots&0\\\vdots&\vdots&&\vdots\\0&0&\cdots&\lambda_n\end{bmatrix}$$

Let $P=\begin{bmatrix}\vec v_1&\vec v_2&\cdots&\vec v_n\end{bmatrix}$ and $Q=\begin{bmatrix}\lambda_1&0&\cdots&0\\0&\lambda_2&\cdots&0\\\vdots&\vdots&&\vdots\\0&0&\cdots&\lambda_n\end{bmatrix}$.

We have $AP=PQ$.

Since the eigenvectors are linearly independent, $P$ is invertible.

Hence, $A=PQP^{-1}$, and we have proved that $A$ is diagonalizable, since $Q$ is a diagonal matrix.