Why We Need Jordan Canonical Form?

By Z.H. Fu https://fuzihaofzh.github.io/blog/

Jordan Canonical Form is an important form of matrix, because not all matrices have the eigen decomposition. But it is guaranteed that all real matrices have Jordan canonical form decomposition.

This article gives an example to show that when a matrix cannot be diagonalized, it is still similar to some approximate diagonal form. We can construct varieties of approximate diagonal form, but the Jordan canonical form is the simplest.

We have

$$A=\begin{bmatrix} 4 &3 &-4\\ -1 &0 &2\\ 1 &1 &0 \end{bmatrix}$$

We solve the equation $AV=\lambda v$ to get the eigen value and the eigenvectors. We get:

$$\begin{aligned} &\lambda_1 = 2,\ \ v_1=(2, 0, 1)^T \\ &\lambda_2 = 1,\ \ v_2=(-1, 1, 0)^T \end{aligned}$$

We find that the number of eigenvectors is 2. It is less than the dimension of the matrix $A$, which means that the eigen vector space cannot span the whole space of $A$. Thus, we should add more vectors into the eigenvectors to span the whole space. So the prerequisite for the selection of complementary vectors is that it should not in the space spanned by existing vectors.

We select $v_3=(1,2,2)^T$. It is not in the space spanned by $v_1$ and $v_2$. So the decomposition of $A$ can be written as $A=PDP^{-1}$, where

$$P=(v_1,v_2,v_3)=\begin{bmatrix}2 &-1 &1 \\ 0 &1 &2 \\ 1 &0 &2\end{bmatrix},P^{-1}=\begin{bmatrix}2 &2 &-3\\2 &3 &-4\\-1 &-1 &2\end{bmatrix}$$

Since $A=PDP^{-1}\Rightarrow AP=PD\Rightarrow$

$$A\begin{bmatrix}v_1 &v_2 &v_3\end{bmatrix}=\begin{bmatrix}v_1 &v_2 &v_3\end{bmatrix}\begin{bmatrix}2 &&\vdots\\&1&a_3\\&&\vdots\end{bmatrix}$$

$\because Av_3=\begin{bmatrix}2\\3\\3\end{bmatrix}=v_1+v_2+v_3$
$\therefore a_3=(1,1,1)^T$
$\therefore$ we have a decomposition of A as:

$$A=\begin{bmatrix}2 &-1 &1 \\ 0 &1 &2 \\ 1 &0 &2\end{bmatrix}\begin{bmatrix}2&&1\\&1&1\\&&1\end{bmatrix}\begin{bmatrix}2 &2 &-3\\2 &3 &-4\\-1 &-1 &2\end{bmatrix}$$

Note that the Jordan canonical form is

$$A=\begin{bmatrix}-1 &1 &2 \\ 1 &0 &0 \\ 0 &1 &1\end{bmatrix}\begin{bmatrix}1&1&\\&1\\&&2\end{bmatrix}\begin{bmatrix}0 &1 &0\\-1 &-1 &2\\1 &1 &-1\end{bmatrix}$$

Compare our decomposition and the Jordan canonical form, we can get the following conclusion:

1. A matrix can be decomposed even though it have not enough eigenvectors;
2. Among the varieties of decompositions, Jordan is the simplest.