A matrix is diagonalisable iff it has a basis of eigenvectors

Statement

Let $F$ be a field and let $A$ be an $n \times n$ matrix over $F$. Then $A$ is diagonalisable if and only if $F^n$ has a basis of eigenvectors of $A$. If $v_1, \dots, v_n$ is such a basis with $A v_i = \lambda_i v_i$, and $P$ has the columns $v_1, \dots, v_n$, then $P^{-1} A P$ is the diagonal matrix with entries $\lambda_1, \dots, \lambda_n$.

Remarks

The equation $AP = PD$ says column by column that $A v_i = \lambda_i v_i$. Equivalently, $F^n$ is the direct sum of the eigenspaces of $A$. Equivalently, the sum of the geometric multiplicities of the eigenvalues is $n$.

Explanations

L1 human

To diagonalise a matrix, we need $n$ independent eigenvectors. We put them as columns into a matrix $P$. Then $P^{-1} A P$ is diagonal, with the eigenvalues on the diagonal.