Defective matrix

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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.

A 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, 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 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[edit]

Any Jordan block of size 2×2 or larger is defective. For example, the n × n Jordan block,

J = 
\lambda & 1            & \;     & \;  \\
\;        & \lambda    & \ddots & \;  \\
\;        & \;           & \ddots & 1   \\
\;        & \;           & \;     & \lambda       

has an eigenvalue, λ, with algebraic multiplicity n, but only one distinct eigenvector,

v = \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}.


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).

See also[edit]


  1. ^ a b Golub & Van Loan (1996, p. 316)