diagonalizable

$A = PDP^{-1}$ for some invertible $P$ and diagonal $D$ ; equivalent to having $n$ independent eigenvectors.

An $n\times n$ matrix $A$ is diagonalizable if $A=PDP^{-1}$ for some invertible $P$ and diagonal $D$ . Building $P$ 's columns from $n$ linearly independent eigenvectors $\mathbf v_1,\dots,\mathbf v_n$ and $D$ 's diagonal from the matching eigenvalues makes $AP=PD$ automatically — so $A$ is diagonalizable if and only if $A$ has a full eigenbasis of $n$ independent eigenvectors, not merely one eigenvector per distinct eigenvalue.