# Find a Nonsingular matrix in Jordan Form

Let $$A= \begin{pmatrix} 0 & 0 & 0\\ 1 & 0 & 0\\ 0 & 1 & 1\\ \end {pmatrix}$$

Find a nonsingular matrix $P$ such that $P^{-1}AP$ is in Jordan form.

The course I am taking uses the textbook "Matrices and Linear Transformation" by Cullen.

The example in the book explains how to find $P$ if I know the characteristic polynomial of $A$.

When I tried to find the characteristic polynomial of this matrix, I got TWO eigenvalues: 0 and 1.

According to the example, I need to first find the matrix J which A is similar to.

Theorem 5.12 in my textbook states:If $A \in F_{n\times n}$ has characteristic polynomial $c(x)=\det(xI-A)=\prod^{r}_{i=1}(x-\lambda_{i})^{s_{i}}$ then $A$ is similar to a matrix $J$ with the $\lambda_{i}$ on the diagonal, zeros and ones on the subdiagonal, and zeros elsewhere.

Am I correct in saying $$J= \begin{pmatrix} 0 & 0 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1\\ \end{pmatrix}$$

Yes, you're correct. The Jordan blocks in your case are $$J_2(0) = \pmatrix{0 & 0\\ 1 & 0} \\ J_1(1) = \pmatrix{1}.$$ You need to look at Theorem 5.13 in your book.