In linear algebra, a defective matrix is a square matrix that does not have a complete basis of eigenvectors, and is therefore not diagonalizable. In particular, an ${\displaystyle n\times n}$ matrix is defective if and only if it does not have ${\displaystyle n}$ linearly independent eigenvectors.^[1] A complete basis is formed by augmenting the eigenvectors with generalized eigenvectors, which are necessary for solving defective systems of ordinary differential equations and other problems.

An ${\displaystyle n\times n}$ defective matrix always has fewer than ${\displaystyle n}$ distinct eigenvalues, since distinct eigenvalues always have linearly independent eigenvectors. In particular, a defective matrix has one or more eigenvalues ${\displaystyle \lambda }$ with algebraic multiplicity ${\displaystyle m>1}$ (that is, they are multiple roots of the characteristic polynomial), but fewer than ${\displaystyle m}$ linearly independent eigenvectors associated with ${\displaystyle \lambda }$. If the algebraic multiplicity of ${\displaystyle \lambda }$ exceeds its geometric multiplicity (that is, the number of linearly independent eigenvectors associated with ${\displaystyle \lambda }$), then ${\displaystyle \lambda }$ is said to be a defective eigenvalue.^[1] However, every eigenvalue with algebraic multiplicity ${\displaystyle m}$ always has ${\displaystyle m}$ linearly independent generalized eigenvectors.

A real symmetric matrix and more generally a Hermitian matrix, and a unitary matrix, is never defective; more generally, a normal matrix (which includes Hermitian and unitary matrices as special cases) is never defective.

Jordan block

Any nontrivial Jordan block of size ${\displaystyle 2\times 2}$ or larger (that is, not completely diagonal) is defective. (A diagonal matrix is a special case of the Jordan normal form with all trivial Jordan blocks of size ${\displaystyle 1\times 1}$ and is not defective.) For example, the ${\displaystyle n\times n}$ Jordan block

${\displaystyle J={\begin{bmatrix}\lambda &1&\;&\;\\\;&\lambda &\ddots &\;\\\;&\;&\ddots &1\\\;&\;&\;&\lambda \end{bmatrix)),}$

has an eigenvalue, ${\displaystyle \lambda }$ with algebraic multiplicity ${\displaystyle n}$ (or greater if there are other Jordan blocks with the same eigenvalue), but only one distinct eigenvector ${\displaystyle Jv_{1}=\lambda v_{1))$, where ${\displaystyle v_{1}={\begin{bmatrix}1\\0\\\vdots \\0\end{bmatrix)).}$ The other canonical basis vectors ${\displaystyle v_{2}={\begin{bmatrix}0\\1\\\vdots \\0\end{bmatrix)),~\ldots ,~v_{n}={\begin{bmatrix}0\\0\\\vdots \\1\end{bmatrix))}$ form a chain of generalized eigenvectors such that ${\displaystyle Jv_{k}=\lambda v_{k}+v_{k-1))$ for ${\displaystyle k=2,\ldots ,n}$.

Any defective matrix has a nontrivial Jordan normal form, which is as close as one can come to diagonalization of such a matrix.

Example

A simple example of a defective matrix is

${\displaystyle {\begin{bmatrix}3&1\\0&3\end{bmatrix)),}$

which has a double eigenvalue of 3 but only one distinct eigenvector

${\displaystyle {\begin{bmatrix}1\\0\end{bmatrix))}$

(and constant multiples thereof).