## Subsection9.2.6Jordan Canonical Form

###### Homework9.2.6.1.

Compute the eigenvalues of $k \times k$ matrix

$$J_k( \mu ) = \left( \begin{array}{c c c c c c } \mu \amp 1 \amp 0 \amp \cdots \amp 0 \amp 0 \\ 0 \amp \mu \amp 1 \amp \ddots \amp 0 \amp 0 \\ \vdots \amp \ddots \amp \ddots \amp \ddots \amp \vdots \amp \vdots \\ 0 \amp 0 \amp 0 \amp \ddots \amp \mu \amp 1 \\ 0 \amp 0 \amp 0 \amp \cdots \amp0 \amp \mu \end{array} \right) \label{chapter09-jordan-block-eqn}\tag{9.2.1}$$

where $k \gt 1 \text{.}$ For each eigenvalue compute a basis for the subspace of its eigenvectors (including the zero vector to make it a subspace).

Hint
• How many linearly independent columns does $\lambda I - J_k( \mu )$ have?

• What does this say about the dimension of the null space $\Null( \lambda I - J_k( \mu ) ) \text{?}$

• You should be able to find eigenvectors by examination.

Solution

Since the matrix is upper triangular and all entries on its diagonal equal $\mu \text{.}$ Now,

\begin{equation*} \mu I - J_k( \mu)= \left( \begin{array}{c c c c c c } 0 \amp 1 \amp 0 \amp \cdots \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp \ddots \amp 0 \amp 0 \\ \vdots \amp \ddots \amp \ddots \amp \ddots \amp \vdots \amp \vdots \\ 0 \amp 0 \amp 0 \amp \ddots \amp 0 \amp 1 \\ 0 \amp 0 \amp 0 \amp \cdots \amp0 \amp 0 \end{array} \right) \end{equation*}

has $m-1$ linearly independent columns and hence its nullspace is one dimensional: $\dim( \Null( \mu I - J_k( \mu) ) ) = 1 \text{.}$ So, we are looking for one vector in the basis of $\Null( \mu I - J_k( \mu) ) \text{.}$ By examination, $J_k( \mu ) e_0 = \mu e_0$ and hence $e_0$ is an eigenvector associated with the only eigenvalue $\mu \text{.}$

The matrix in (9.2.1) is known as a Jordan block.

The point of the last exercise is to show that if $A$ has an eigenvalue of algebraic multiplicity $k \text{,}$ then it does not necessarily have $k$ linearly independent eigenvectors. That, in turn, means there are matrices that do not have a full set of eigenvectors. We conclude that there are matrices that are not diagonalizable. We call such matrices defective.

###### Definition9.2.6.1. Defective matrix.

A matrix $A \in \Cmxm$ that does not have $m$ linearly independent eigenvectors is said to be defective.

###### Definition9.2.6.3. Geometric multiplicity.

Let $\lambda \in \Lambda( A ) \text{.}$ Then the geometric multiplicity of $\lambda$ is defined to be the dimension of ${\cal E}_\lambda( A )$ defined by

\begin{equation*} {\cal E}_\lambda( A ) = \{ x \in \C^m \vert A x = \lambda x \}. \end{equation*}

In other words, the geometric multiplicity of $\lambda$ equals the number of linearly independent eigenvectors that are associated with $\lambda \text{.}$

###### Homework9.2.6.2.

Let $A \in \Cmxm$ have the form

\begin{equation*} A = \left( \begin{array}{c | c} A_{00} \amp 0 \\ \hline 0 \amp A_{11} \end{array} \right) \end{equation*}

where $A_{00}$ and $A_{11}$ are square. Show that

• If $( \lambda, x )$ is an eigenpair of $A_{00}$ then $( \lambda, \left( \begin{array}{c} x \\ 0 \end{array} \right) )$ is an eigenpair of $A \text{.}$

• If $( \mu, y )$ is an eigenpair of $A_{11}$ then $( \mu, \left( \begin{array}{c} 0 \\ y \end{array} \right) )$ is an eigenpair of $A \text{.}$

• If $( \lambda, \left( \begin{array}{c} x \\ y \end{array} \right) )$ is an eigenpair of $A$ then $( \lambda, x )$ is an eigenpair of $A_{00}$ and $( \lambda, y )$ is an eigenpair of $A_{11} \text{.}$

• $\Lambda( A ) = \Lambda( A_{00} ) \cup \Lambda( A_{11} ) \text{.}$

Solution

Let $A \in \Cmxm$ have the form

\begin{equation*} A = \left( \begin{array}{c | c} A_{00} \amp 0 \\ \hline 0 \amp A_{11} \end{array} \right) \end{equation*}

where $A_{00}$ and $A_{11}$ are square. Show that

• If $( \lambda, x )$ is an eigenpair of $A_{00}$ then $( \lambda, \left( \begin{array}{c} x \\ 0 \end{array} \right))$ is an eigenpair of $A \text{.}$

\begin{equation*} \left( \begin{array}{c | c} A_{00} \amp 0 \\ \hline 0 \amp A_{11} \end{array} \right) \left( \begin{array}{c} x \\ 0 \end{array} \right) = \left( \begin{array}{c} A_{00} x \\ 0 \end{array} \right) = \left( \begin{array}{c} \lambda x \\ 0 \end{array} \right) = \lambda \left( \begin{array}{c} x \\ 0 \end{array} \right) . \end{equation*}
• If $( \mu, y )$ is an eigenpair of $A_{11}$ then $( \mu, \left( \begin{array}{c} 0 \\ y \end{array} \right) )$ is an eigenpair of $A \text{.}$

\begin{equation*} \left( \begin{array}{c | c} A_{00} \amp 0 \\ \hline 0 \amp A_{11} \end{array} \right) \left( \begin{array}{c} 0 \\ y \end{array} \right) = \left( \begin{array}{c c} 0 \amp A_{11} y \end{array} \right) = \left( \begin{array}{c c} 0 \amp \mu y \end{array} \right) = \mu \left( \begin{array}{c} 0 \\ y \end{array} \right) . \end{equation*}
• $\left( \begin{array}{c | c} A_{00} \amp 0 \\ \hline 0 \amp A_{11} \end{array} \right) \left( \begin{array}{c} x \\ \hline y \end{array} \right) = \lambda \left( \begin{array}{c} x \\ \hline y \end{array} \right)$ implies that

\begin{equation*} \left( \begin{array}{c c} A_{00} x \\ \hline A_{11} y \end{array} \right) = \left( \begin{array}{c c} \lambda \\ \hline \lambda y \end{array} \right), \end{equation*}

and hence $A_{00} x = \lambda x$ and $A_{11} y = \lambda y \text{.}$

• $\Lambda( A ) = \Lambda( A_{00} ) \cup \Lambda( A_{11} ) \text{.}$

This follows from the first three parts of this problem.

This last homework naturally extends to

\begin{equation*} A = \left( \begin{array}{c | c | c | c } A_{00} \amp 0 \amp \cdots \amp 0\\ \hline 0 \amp A_{11} \amp \cdots \amp 0 \\ \hline \vdots \amp \vdots \amp \ddots \amp \vdots \\ \hline 0 \amp 0 \amp \cdots \amp A_{kk} \end{array} \right) \end{equation*}

The following is a classic result in linear algebra theory that characterizes the relationship between of a matrix and its eigenvectors:

For our discussion, the sizes of the Jordan blocks $J_{m_i}( \lambda_i )$ are not particularly important. Indeed, this decomposition, known as the Jordan Canonical Form of matrix $A \text{,}$ is not particularly interesting in practice. It is extremely sensitive to perturbation: even the smallest random change to a matrix will make it diagonalizable. As a result, there is no practical mathematical software library or tool that computes it. For this reason, we don't give its proof and don't discuss it further.