# Understanding the Jordan form via this example

## The Question:

I do not understand why the Jordan Form of the matrix: $A:=\begin{pmatrix} 1 & 1 \\ -1 & 3 \end{pmatrix}$ is: $J:=\begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$?

## Here is what I have done so far:

The characteristic polynomial of $A$ is given by: $\phi(t)=t^2-4t+4$ and $v:=(1,1)$ is an eigenvector of $A$ corresponding to the eigenvalue $2$.

However my issue is that $(A-2I)v=0$, but apparently $v$ is part of a length $2$ cycle of generalised eigenvectors, how do I find the other member of this cycle of eigenvectors?

you can work backwards.

if $A$ is similar to $J,$ then there is a nonsingular matrix $P$ made up columns $u, v$ so that $$A[u,v]= [u,v]J=[u,v]\pmatrix{2&1\\0&2}$$ this is equivalent to two equations $$(A-2I)u = 0, (A-2I)v = u.$$ the vector $v$ is called the generalized eigenvector corresponding to the eigenvalue $2.$ the chain $u, v$ is called the jordan chain of length $2$ corresponding to the eigenvalue $2.$

we have $$A - 2I= \pmatrix{-1&1\\-1&1}, u = \pmatrix{1\\1}$$ to find the generalized eigenvector $v$, we will solve the system $$\pmatrix{-1&1&|&1\\-1&1&|&1}\to \pmatrix{-1&1&|&1\\0&0&|&0}$$ so $$v = \pmatrix{0\\1}$$ will do and $$\pmatrix{1&1\\-1&3}\pmatrix{1&0\\1&1}= \pmatrix{1&0\\1&1}\pmatrix{2&1\\0&2} \to \\\pmatrix{1&1\\-1&3}=\pmatrix{1&0\\1&1}\pmatrix{2&1\\0&2}\pmatrix{1&-1\\0&1}$$

• Perfect answer you've finally demystified the Jordan form for me entirely :D – AIM_BLB May 3 '15 at 15:23

Alternative explanation

We have the matrix $A$, that we want to put into Jordan form $J$. $$A= \begin{bmatrix}1&1\\-1&3\end{bmatrix}$$

What are the eigenvalues? $$(1-\lambda)(3-\lambda)+1=0\iff \lambda^2-4\lambda+4=0\iff (\lambda-2)^2=0$$ Hence we have repeated eigenvalues at $2$.

A Jordan normal form of size $2\times 2$ takes the following forms:

$$\begin{bmatrix}\lambda&0\\0&\mu\end{bmatrix},\begin{bmatrix}\lambda&0\\0&\lambda\end{bmatrix},\begin{bmatrix}\lambda&1\\0&\lambda\end{bmatrix}$$

Where $\lambda\ne \mu$ and $\lambda,\mu$ are eigenvalues. Hence we have either:

$$\begin{bmatrix}2&0\\0&2\end{bmatrix}\text{, or }\begin{bmatrix}2&1\\0&2\end{bmatrix}$$

Can $A$ be diagonalized though? The answer is no. If it were diagonalizable, we would obtain the first of the two options above, and hence we must have the second:

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

• Thanks for the solution :) – AIM_BLB May 3 '15 at 15:24

Since $\ker(A-2I)\varsubsetneq\ker(A-2I)^2 \cong\mathbf R^2$, all you have to do is solve for the equation $(A-2I)\cdot u=v$, so that $$Av=2v, \quad Au=2u+v,$$ hence the matrix of $A$ in the $(v,u)$ basis will be $\,\begin{bmatrix} 2&1\\0&2\end{bmatrix}$.

• Thanks for the solution Bernard :) – AIM_BLB May 3 '15 at 15:25

Just solve the equation $$Au = 2u+v.$$

Then, the matrix in the base $(u,v)$ will be $J^T$.

• Thanks for the solution mookid! – AIM_BLB May 3 '15 at 15:24