Lab 8. Jordan Canonical Form: Independent Study
This lab accompanies Chapter 8: Jordan Canonical Form.
The goal is to understand what happens when diagonalization fails:
- repeated eigenvalues may not give enough eigenvectors;
- generalized eigenvectors repair the missing directions;
- Jordan chains create Jordan blocks;
- powers and exponentials of Jordan blocks have polynomial correction terms;
- kernel dimensions reveal Jordan block sizes.
This is an independent-study lab. Each main question includes a worked solution and a similar practice question.
Python practice notebook
You may use the Jupyter notebook version for longer Python practice:
Interactive lab
Study guide and worked questions
Question 1. A matrix that is not diagonalizable
Let
\[ A=\begin{bmatrix}2&1\\0&2\end{bmatrix}. \]
Show that \(A\) is not diagonalizable.
Solution
The characteristic polynomial is
\[ \det(A-\lambda I)=(2-\lambda)^2. \]
So the only eigenvalue is \(\lambda=2\), with algebraic multiplicity \(2\).
Now
\[ A-2I=\begin{bmatrix}0&1\\0&0\end{bmatrix}. \]
The eigenspace is
\[ E_2=\ker(A-2I)=\operatorname{span}\left\{\begin{bmatrix}1\\0\end{bmatrix}\right\}. \]
It has dimension \(1\), not \(2\). Therefore \(A\) does not have enough eigenvectors to form a basis and is not diagonalizable.
Similar practice
Show that
\[ B=\begin{bmatrix}5&1\\0&5\end{bmatrix} \]
is not diagonalizable.
Answer
The only eigenvalue is \(5\). The eigenspace is
\[ \ker(B-5I)=\operatorname{span}\left\{\begin{bmatrix}1\\0\end{bmatrix}\right\}, \]
which is one-dimensional. Hence \(B\) is not diagonalizable.
Question 2. Find a Jordan chain
For
\[ A=\begin{bmatrix}2&1\\0&2\end{bmatrix}, \]
find a Jordan chain.
Solution
Let
\[ v_1=e_1=\begin{bmatrix}1\\0\end{bmatrix}, \qquad v_2=e_2=\begin{bmatrix}0\\1\end{bmatrix}. \]
Then
\[ (A-2I)v_1=0, \qquad (A-2I)v_2=v_1. \]
Thus \(v_1,v_2\) is a Jordan chain for the eigenvalue \(2\).
Similar practice
Find a Jordan chain for
\[ B=\begin{bmatrix}5&1\\0&5\end{bmatrix}. \]
Answer
Use \(v_1=e_1\) and \(v_2=e_2\). Then
\[ (B-5I)v_1=0, \qquad (B-5I)v_2=v_1. \]
Question 3. Powers of a Jordan block
Let
\[ J=\begin{bmatrix}3&1\\0&3\end{bmatrix}. \]
Compute \(J^m\).
Solution
Write
\[ J=3I+N, \qquad N=\begin{bmatrix}0&1\\0&0\end{bmatrix}, \qquad N^2=0. \]
Then
\[ J^m=(3I+N)^m=3^mI+m3^{m-1}N. \]
Therefore
\[ J^m=\begin{bmatrix}3^m&m3^{m-1}\\0&3^m\end{bmatrix}. \]
Similar practice
For
\[ K=\begin{bmatrix}2&1\\0&2\end{bmatrix}, \]
compute \(K^m\).
Answer
\[ K^m=\begin{bmatrix}2^m&m2^{m-1}\\0&2^m\end{bmatrix}. \]
Question 4. Kernel dimensions and Jordan block sizes
Suppose for one eigenvalue \(\lambda\) we have
\[ d_1=2, \qquad d_2=4, \qquad d_3=5, \]
where \(d_j=\dim\ker((A-\lambda I)^j)\). Find the Jordan block sizes.
Solution
Let \(d_0=0\). Then
\[ d_1-d_0=2, \qquad d_2-d_1=2, \qquad d_3-d_2=1. \]
So there are two blocks of size at least \(1\), two blocks of size at least \(2\), and one block of size at least \(3\). Hence the Jordan block sizes are
\[ 3 \quad\text{and}\quad 2. \]
Similar practice
Suppose
\[ d_1=3, \qquad d_2=5, \qquad d_3=6. \]
Find the Jordan block sizes.
Answer
The differences are
\[ 3,2,1. \]
Therefore the block sizes are
\[ 3,2,1. \]
Question 5. Matrix exponential of a Jordan block
Let
\[ J=\begin{bmatrix}\lambda&1\\0&\lambda\end{bmatrix}. \]
Find \(e^{tJ}\).
Solution
Write \(J=\lambda I+N\), where
\[ N=\begin{bmatrix}0&1\\0&0\end{bmatrix}, \qquad N^2=0. \]
Then
\[ e^{tJ}=e^{\lambda t}e^{tN}=e^{\lambda t}(I+tN). \]
Thus
\[ e^{tJ}=e^{\lambda t} \begin{bmatrix}1&t\\0&1\end{bmatrix}. \]
Similar practice
For
\[ J=\begin{bmatrix}2&1\\0&2\end{bmatrix}, \]
find \(e^{tJ}\).
Answer
\[ e^{tJ}=e^{2t}\begin{bmatrix}1&t\\0&1\end{bmatrix}. \]
Short reflection
In your own words, explain this sentence:
A Jordan block is an eigenvalue plus a nilpotent correction.
A good answer should mention the formula
\[ J_{\lambda,k}=\lambda I+N_k \]
and explain how the nilpotent part creates generalized eigenvector chains.