Defective matrix

Template:Short description Template:Use American English 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

$J=\left[\begin{array}{cccc}\lambda & 1& & \\ & \lambda & \ddots & \\ & & \ddots & 1\\ & & & \lambda \end{array}\right],$

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 $v_1=\left[\begin{array}{c}1\\ 0\\ ⋮\\ 0\end{array}\right].$ The other canonical basis vectors $v_2=\left[\begin{array}{c}0\\ 1\\ ⋮\\ 0\end{array}\right],\dots ,v_n=\left[\begin{array}{c}0\\ 0\\ ⋮\\ 1\end{array}\right]$ form a chain of generalized eigenvectors such that $J{v}_k=\lambda v_k+v_{k-1}$ for $k=2,\dots ,n$.

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

$\left[\begin{array}{cc}3& 1\\ 0& 3\end{array}\right],$

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

$\left[\begin{array}{c}1\\ 0\end{array}\right]$

(and constant multiples thereof).

See also

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References

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