Jordan Form Example

A simple walked through example.

Consider the matrix

\[A = \begin{bmatrix}0 & 4 & 3 \\ 0 & 20 & 16 \\ 0 & -25 & -20\end{bmatrix}\]

and we wish to find the Jordan form \(A = V J V^{-1}\) of this matrix.

Step 1: Find Eigenvalues:

First we determine the eigenvalues:

\(det(A - \lambda I) = 0\) and this has three repeated solutions \(\lambda = 0\).

Thus there is one unique eigenvalue \(\lambda_1 = 0\), with algebraic multiplicity \(m_1 = 3\).

Step 2: Determine the form of the Jordan Block:

We use the geometric multiplicity, which is defined as the dimension of the eigen space:

\[q_i = \text{dim} \mathcal{N}(A-\lambda I)\]

In this case

\[\mathcal N (A - 0 I ) = \text{span}\left(\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}\right)\]

and so \(q_1 = 1\), and so there is only one type of Jordan block that we can have:

\[J = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\]

normally, you will have a set of possible Jordan blocks, and which one it is depends on the index of the eigenvalue.

Step 3: Find Generalized Eigenspaces:

In this case, we have a single eigenvector of rank 1, and so we will have a single chain of generalized eigenvectors. We start by finding the generalied eigenspaces:

\[\mathcal{N}(A-\lambda I) = \mathcal{N}\begin{bmatrix}0 & 4 & 3 \\ 0 & 20 & 16 \\ 0 & -25 & -20\end{bmatrix} = \text{span} \left(\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}\right)\] \[\mathcal{N}\left((A-\lambda I)^2\right) = \mathcal{N}\begin{bmatrix}0 & -5 & 4 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix} = \text{span} \left(\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 4 \\ 5\end{bmatrix}\right)\] \[\mathcal{N}\left((A-\lambda I)^3\right) = \mathcal{N}\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix} = \text{span} \left(\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}\right)\]

Step 4: Find the Generalised Eigenvectors:

and now we must work backwards - we must choose a \(v^3, v^2, v^1\) such that

\[v^3 \in \mathcal{N}\left((A-\lambda I)^3\right) \text{ and } v^3 \not \in \mathcal{N}\left((A-\lambda I)^2\right)\]

say we choose

\[v^3 = \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}\]

Then we can determine \(v^2\):

\[v^2 = (A - \lambda I) v^3 = \begin{bmatrix}3 \\ -16 \\ 20\end{bmatrix}\]

and \(v^1\):

\[v^1 = (A - \lambda I) v^2 = \begin{bmatrix}4 \\ 0 \\ 0\end{bmatrix}\]

which gives us the basis:

\[V = \begin{bmatrix} v^1 & v^2 & v^3 \end{bmatrix} = \begin{bmatrix} 4 & 3 & 0\\ 0 & -16 & 0 \\ 0 & 20 & 1 \end{bmatrix}\]

Solution:

So the Jordan form is:

\[J = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \quad V =\begin{bmatrix} 4 & 3 & 0\\ 0 & -16 & 0 \\ 0 & 20 & 1 \end{bmatrix}\]

and you can verify that \(A = V J V^{-1}\).

Note, a different choice for \(v^3\) will give a different matrix \(V\), but the Jordan form \(J\) is the same, and the decomposition is also still valid.

Also important to note, you can’t rescale generalised eigenvectors in quite the same way as with normal eigenvectors. If you rescale \(v^3\) for instance, you have to recompute all the vectors in the chain so that they match up.