Non-diagonalizable matrix; one lacking a basis of eigenvectors

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 $n\times n$ matrix is defective if and only if it does not have $n$ 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 $n\times n$ 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 $\lambda$ with algebraic multiplicity $m>1$ (that is, they are multiple roots of the characteristic polynomial), but fewer than $m$ linearly independent eigenvectors associated with $\lambda$. If the algebraic multiplicity of $\lambda$ exceeds its geometric multiplicity (that is, the number of linearly independent eigenvectors associated with $\lambda$), then $\lambda$ is said to be a **defective eigenvalue**.^{[1]} However, every eigenvalue with algebraic multiplicity $m$ always has $m$ 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 $2\times 2$ 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).