= Solution
A square <matrix> $A$ is a <diagonalizable matrix> over a specified field if some invertible <matrix> $P$ makes $P^{-1}AP$ diagonal. Equivalently, the underlying <vector space> has a <basis> of <eigenvectors> of $A$.
For example,
$$
\boxed{\begin{pmatrix}1&0\\0&2\end{pmatrix}\ \text{is diagonalizable},\qquad
\begin{pmatrix}1&1\\0&1\end{pmatrix}\ \text{is not diagonalizable}}.
$$
The first is already diagonal. The second has only the <eigenvalue> one, of algebraic multiplicity two, but its <eigenspace> consists of $(x,0)^T$ and has dimension one. It cannot supply a two-vector <eigenbasis>.
Back to article page
