# Find a nonsingular matrix P given that A is similar to a Jordan matrix

Given ${\bf A}$ is similar to a Jordan matrix find a nonsingular matrix $\bf P$ such that ${\bf P}^{-1}{\bf AP}={\bf J}$

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

I have worked out $${\bf J}= \begin{bmatrix} 0&0&0\\ 1&0&0\\ 0&0&1\\ \end{bmatrix}$$

The textbook I am using shows an example where ${\bf A}$ has one eigenvalue. I am unsure how to apply this example to the question I have.

I am using "Matrices and Linear Transformations" by Cullen. The example I was looking at is on page204.

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$A$ has three eigenvalues, or at least one at $1$ and one at $0$ of multiplicity 1. This tells you the general form of the Jordan form. Looking at $\ker A$ gives the $(0,1,-1)^T$ vector, $\ker (A-I)$ gives $(0,0,1)^T$, and looking at $\ker A^2$ yields the $(1,0,-1)^T$ vector. After that, it is a matter of permuting these vectors to get $P$.
Try $P=\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ -1 & -1 & 1 \end{bmatrix}$. Then $P^{-1} A P = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$, which is a Jordan form.