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In a question set in my linear algebra course, I'm asked the following:

Find $P$ such that $P^{-1}AP=J$, where $$A = \begin{pmatrix}6&5&-2&-3\\ -3&-1&3&3\\ 2&1&-2&-3\\-1&1&5&5\end{pmatrix}$$ and $$J = \begin{pmatrix}2&1&0&0\\ 0&2&0&0\\ 0&0&2&1\\0&0&0&2\end{pmatrix}$$

I can see that there's only one eigenvalue and two eigenvectors associated to it, but I don't have a clue as to how to proceed. While searching online, I didn't find any reasonable way to find $P$. Please give me a guideline as to how I should proceed. Thanks in advance.

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  • $\begingroup$ There is an easy way. Compute directly one of several $P$ with $AP=PJ$, using matrix multiplication with an unknown $P$. Then choose an invertible one from these. More systematically, use this duplicate. One possibility is $\begin{pmatrix} 2 & 1 & -4 & -1\cr 0 & 0 & 3 & 0\cr -2 & 1 & -2 & 0 \cr 4 & 0 & 1 & 0\end{pmatrix}$. $\endgroup$ – Dietrich Burde May 25 at 18:58
  • $\begingroup$ What would you consider “reasonable?” The standard methods that you’ve likely found in your searches don’t require a tremendous amount of work here, and knowing the JNF ahead of time saves you a bit of work. $\endgroup$ – amd May 26 at 0:56
  • $\begingroup$ You can a little simplify problem if you notice that $J=2I+N$, where $N$ is a simple nilpotent matrix. Then $P(2I+N)=AP$ and conseqently $PN=(A-2I)P$ $\endgroup$ – Widawensen May 26 at 13:24
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You want $P^{-1}AP=J$, which is equivalent to $P^{-1}(A-2I)P=J-2I$ or $$(A-2I)P=P(J-2I)$$ (assuming that $P$ is regular.)

We have $$J-2I= \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix} $$

Let us denote the columns of $P$ by $\vec p_1, \dots, \vec p_4$. Then you get \begin{align*} (A-2I)P&=PJ\\ (A-2I)\begin{pmatrix}\vec p_1&\vec p_2&\vec p_3&\vec p_4\end{pmatrix}&= \begin{pmatrix}\vec p_1&\vec p_2&\vec p_3&\vec p_4\end{pmatrix} J\\ \begin{pmatrix}(A-2I)\vec p_1&(A-2I)\vec p_2&(A-2I)\vec p_3&(A-2I)\vec p_4\end{pmatrix}&= \begin{pmatrix} \vec0 &\vec p_1&\vec 0&\vec p_3\end{pmatrix} \end{align*} Let us have a look at $\vec p_1$ and $\vec p_2$. (The situation for $\vec p_3$ and $\vec p_4$ is similar.) We got \begin{align*} (A-2I)\vec p_1&=\vec p_2\\ (A-2I)\vec p_2&=\vec 0 \end{align*} From this we see that $\vec p_2$ is an eigenvector. Moreover, we get $$(A-2I)^2\vec p_1=\vec0.$$ Exactly in the same way we get $(A-2I)\vec p_3=\vec p_4$ and $(A-2I)^2\vec p_3=\vec0$.

You can notice that $(J-2I)^2=0$ and, consequently, $(A-2I)^2=0$. This means that $(A-2I)^2\vec x=\vec 0$ is true for any vector. However, for $\vec p_1$ and $\vec p_3$ we want to choose such vectors which are linearly independent and, moreover $(A-2I)\vec p_1$ and $(A-2I)\vec p_3$. (Since to get invertible matrix we need $\vec p_2$ and $\vec p_4$ to be linearly independent.)

It is certainly not that difficult to find two vectors such that $(A-2I)\vec p_1$ and $(A-2I)\vec p_3$ are linearly independent. We can then find the two remaining columns using $\vec p_2=(A-2I)\vec p_1$ and $\vec p_4=(A-2I)\vec p_3$.

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