Diagonalizable operator

A linear operator $T:V\to V$ on a finite-dimensional vector space $V$ over a field $K$ is diagonalizable over $K$ if $V$ has a basis consisting of eigenvectors of $T$ .

Equivalent characterizations

Equivalently, $T$ is diagonalizable iff its matrix representation in some basis is diagonal.

The following are also equivalent:

$T$ is diagonalizable.
$V = \bigoplus_{\lambda} E_\lambda$ where $E_\lambda$ is the eigenspace for eigenvalue $\lambda$ .
The sum of geometric multiplicities equals $\dim V$ .
The minimal polynomial of $T$ splits into distinct linear factors.

Criteria

If the characteristic polynomial splits over $K$ and has distinct roots, then $T$ is diagonalizable. More generally, when the characteristic polynomial splits over $K$ , $T$ is diagonalizable if and only if each eigenvalue's geometric multiplicity equals its algebraic multiplicity.

Examples

Any operator with $n$ distinct eigenvalues (where $\dim V = n$ ).
Self-adjoint operators on finite-dimensional real or complex inner product spaces.
Projections.

Non-example

The matrix $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ is not diagonalizable (its eigenspace is only one-dimensional).