# Defective matrix

Short description: 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 × 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 × 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.[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 $\displaystyle{ 2 \times 2 }$ 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 n (or greater if there are other Jordan blocks with the same eigenvalue), but only one distinct eigenvector $\displaystyle{ J v_1 = \lambda v_1 }$, 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 $\displaystyle{ J v_k = \lambda v_k + v_{k-1} }$ 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).

• Jordan normal form – Form of a matrix indicating its eigenvalues and their algebraic multiplicities

## Notes

1. (Golub Van Loan)