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.

  • $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.