Defective matrix

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. 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. 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$$ (or greater if there are other Jordan blocks with the same eigenvalue), but only one distinct eigenvector $$ J v_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}$$ form a chain of generalized eigenvectors such that $$J v_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).