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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.

Please help.

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1 Answer 1

up vote 2 down vote accepted

$A$ 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.

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That second entry on the first row doesn't look compatible with Jordan form. –  Erick Wong Jul 15 '12 at 19:51
    
Oops, good catch. I'll fix it. –  copper.hat Jul 15 '12 at 20:07
    
There is a homework tag.. –  user2468 Jul 15 '12 at 21:00
    
How do I add a hint? –  copper.hat Jul 15 '12 at 21:30
1  
Click edit to see the source code of any answer by Marvis But seriously, we shouldn't be giving all the details of the solution.. –  user2468 Jul 16 '12 at 3:28

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