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 n × n matrix is defective if and only if it does not have n linearly independent eigenvectors. 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 n × n defective matrix always has fewer than n distinct eigenvalues, since distinct eigenvalues always have linearly independent eigenvectors. In particular, a defective matrix has one or more eigenvalues λ with algebraic multiplicity m > 1 (that is, they are multiple roots of the characteristic polynomial), but fewer than m linearly independent eigenvectors associated with λ. If the algebraic multiplicity of λ exceeds its geometric multiplicity (that is, the number of linearly independent eigenvectors associated with λ), then λ is said to be a defective eigenvalue. However, every eigenvalue with algebraic multiplicity m always has m linearly independent generalized eigenvectors.

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

## Jordan block

Any nontrivial Jordan block of size $2\times 2$ 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 $1\times 1$ and is not defective.) For example, the $n\times n$ Jordan block

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

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

${\begin{bmatrix}3&1\\0&3\end{bmatrix)),$ which has a double eigenvalue of 3 but only one distinct eigenvector

${\begin{bmatrix}1\\0\end{bmatrix))$ (and constant multiples thereof).