Generalized Eigenvectors and Defective Matrices

When a matrix $A$ has repeated eigenvalues with insufficient eigenvectors, generalized eigenvectors and Jordan chains provide the tools to construct a complete solution.

Defective Matrices

Definition5.4Defective matrix

A matrix $A$ is defective if it does not have $n$ linearly independent eigenvectors. This occurs when the geometric multiplicity of some eigenvalue is strictly less than its algebraic multiplicity. For such matrices, Jordan chains replace the missing eigenvectors.

Definition5.5Generalized eigenvector

A generalized eigenvector of rank $k$ for eigenvalue $\lambda$ is a nonzero vector $\mathbf{v}$ satisfying $(A - \lambda I)^k \mathbf{v} = \mathbf{0}$ but $(A - \lambda I)^{k-1}\mathbf{v} \neq \mathbf{0}$ . A Jordan chain of length $k$ is $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k$ where $(A-\lambda I)\mathbf{v}_1 = \mathbf{0}$ and $(A-\lambda I)\mathbf{v}_{j+1} = \mathbf{v}_j$ .

Solutions from Jordan Chains

Theorem5.5Solutions from generalized eigenvectors

If $\lambda$ has algebraic multiplicity $m$ and geometric multiplicity $p < m$ , the $m$ linearly independent solutions corresponding to $\lambda$ involve terms of the form $t^k e^{\lambda t}\mathbf{v}$ where $\mathbf{v}$ are generalized eigenvectors. Specifically, for a Jordan chain $\mathbf{v}_1, \ldots, \mathbf{v}_k$ :

$\mathbf{x}_j(t) = e^{\lambda t}\sum_{i=0}^{j-1}\frac{t^i}{i!}\mathbf{v}_{j-i}, \quad j = 1, \ldots, k.$

ExampleDefective $3 \times 3$ system

Solve $\mathbf{x}' = A\mathbf{x}$ where $A = \begin{pmatrix}2&1&0\\0&2&1\\0&0&2\end{pmatrix}$ (one eigenvalue $\lambda=2$ , multiplicity 3).

Only one eigenvector: $\mathbf{v}_1 = (1,0,0)^T$ . Generalized: $(A-2I)\mathbf{v}_2 = \mathbf{v}_1$ gives $\mathbf{v}_2 = (0,1,0)^T$ . Then $(A-2I)\mathbf{v}_3 = \mathbf{v}_2$ gives $\mathbf{v}_3 = (0,0,1)^T$ .

$\mathbf{x}(t) = e^{2t}\left[c_1\begin{pmatrix}1\\0\\0\end{pmatrix} + c_2\left(\begin{pmatrix}0\\1\\0\end{pmatrix}+t\begin{pmatrix}1\\0\\0\end{pmatrix}\right) + c_3\left(\begin{pmatrix}0\\0\\1\end{pmatrix}+t\begin{pmatrix}0\\1\\0\end{pmatrix}+\frac{t^2}{2}\begin{pmatrix}1\\0\\0\end{pmatrix}\right)\right].$

The Jordan Normal Form

RemarkJordan blocks

Every square matrix is similar to a Jordan normal form $J = \text{diag}(J_1, J_2, \ldots, J_r)$ where each Jordan block is

$J_k = \begin{pmatrix}\lambda_k & 1 & & 0 \\ & \lambda_k & 1 & \\ & & \ddots & 1 \\ 0 & & & \lambda_k\end{pmatrix}.$

The matrix exponential of a Jordan block is $e^{J_k t} = e^{\lambda_k t}\begin{pmatrix}1 & t & t^2/2 & \cdots \\ 0 & 1 & t & \cdots \\ & & \ddots & \\ 0 & & & 1\end{pmatrix}$ .

Example$2 \times 2$ Jordan block example

For $A = \begin{pmatrix}3 & 1\\-1 & 5\end{pmatrix}$ , eigenvalue $\lambda = 4$ (multiplicity 2, geometric multiplicity 1).

Eigenvector: $\mathbf{v}_1 = (1, 1)^T$ . Generalized: $(A-4I)\mathbf{v}_2 = \mathbf{v}_1 \implies \mathbf{v}_2 = (0, 1)^T$ .

$\mathbf{x}(t) = c_1 e^{4t}\begin{pmatrix}1\\1\end{pmatrix} + c_2 e^{4t}\left[\begin{pmatrix}0\\1\end{pmatrix} + t\begin{pmatrix}1\\1\end{pmatrix}\right].$