# Given the Jordan form $J$, find matrix $P$

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.

• 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 $\begin{pmatrix} 2 & 1 & -4 & -1\cr 0 & 0 & 3 & 0\cr -2 & 1 & -2 & 0 \cr 4 & 0 & 1 & 0\end{pmatrix}$. – Dietrich Burde May 25 at 18:58
• 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. – amd May 26 at 0:56
• 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 $PN=(A-2I)P$ – Widawensen May 26 at 13:24

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 $$\vec p_4=(A-2I)\vec p_3$$.