Diagonalizable operator

A linear operator $T: V \to V$ on a finite-dimensional vector space is diagonalizable if there exists a basis of $V$ consisting entirely of eigenvectors of $T$.

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

Characterizations

The following are equivalent:

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

Over algebraically closed fields

If the characteristic polynomial has distinct roots, then $T$ is diagonalizable. More generally, $T$ is diagonalizable iff each eigenvalue’s geometric multiplicity equals its algebraic multiplicity.

Examples

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

Non-example

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