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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} $

Please help.

1 Answers 1

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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.

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    I guess this is now covered by copper.hat's answer on your other question: [click here](http://math.stackexchange.com/questions/171235/find-a-nonsingular-matrix-p-given-that-a-is-similar-to-a-jordan-matrix).2012-07-16