1

Assume matrix

$$A= \begin{bmatrix} -1&0&0&0&0\\ -1&1&-2&0&1\\ -1&0&-1&0&1\\ 0&1&-1&1&0\\ 0&0&0&0&-1 \end{bmatrix} $$

Its Jordan Canonical Form is $$J= \begin{bmatrix} -1&1&0&0&0\\ 0&-1&0&0&0\\ 0&0&-1&0&0\\ 0&0&0&1&1\\ 0&0&0&0&1 \end{bmatrix} $$

I am trying to find a nonsingular $P$, let $P=\begin{bmatrix}\mathbf{p}_1&\mathbf{p}_2&\mathbf{p}_3&\mathbf{p}_4&\mathbf{p}_5\end{bmatrix}$ s.t. $J=P^{-1}AP\Leftrightarrow AP=PJ$. I came up with the Wikipedia article on JCF and I think I need to find the generalized eigenvectors so that $AP=PJ=\begin{bmatrix}-\mathbf{p_1}&\mathbf{p_1}-\mathbf{p_2}&-\mathbf{p_3}&\mathbf{p_4}&\mathbf{p_4}+\mathbf{p_5}\end{bmatrix}$ yielding the systems $$(A+I)\mathbf{p_1}=\mathbf{0}$$ $$(A+I)^2\mathbf{p_2}=\mathbf{0}$$ $$(A+I)\mathbf{p_3}=\mathbf{0}$$ $$(A-I)\mathbf{p_4}=\mathbf{0}$$ $$(A-I)^2\mathbf{p_5}=\mathbf{0}$$

I solved each of these systems making sure that the vectors $\mathbf{p_i}$ I chose are linearly independent. So I chose $$P=\begin{bmatrix}\mathbf{p}_1&\mathbf{p}_2&\mathbf{p}_3&\mathbf{p}_4&\mathbf{p}_5\end{bmatrix}=\begin{bmatrix}1&2&-2&0&0\\1&1&2&0&1\\1&1&2&0&0\\0&0&0&1&1\\1&1&-2&0&0\end{bmatrix}$$ which even though is nonsingular I am not getting $AP=PJ$.

What am I doing wrong?

mgus
  • 1,381
  • are you sure for J? if so you can solve by the given hint – user Mar 12 '18 at 23:31
  • @gimusi I had a typo in matrix A when I typed in SE but now I fixed it. So the JCF is still the same matrix. – mgus Mar 12 '18 at 23:48
  • but how have you determined the structure of J? I've not checked,are you sure about it? – user Mar 12 '18 at 23:52
  • @gimusi As you probably know the JCF of the matrix is the same up to permutations of the Jordan blocks. But I used this specific structure typed here for my further calculations. – mgus Mar 12 '18 at 23:56
  • Yes, I'm asking if you are sure about the blocks. If so you can solve as indicated in the hint! Have you tried? – user Mar 12 '18 at 23:58
  • ok I've checked and it is right! thus what is the problem to find P? – user Mar 13 '18 at 00:09
  • @gimusi Have you read my whole post? Because I think I have already done what you suggested. So, I found $p_1$ then found $p_2$ so that is linearly independent with $p_1$, then $p_3$ linearly independent with all previous ones and so on. In this way, I found a nonsingular $P$ as stated in the post. But that $P$ does not satisfy the equation $J=P^{-1}AP$ and I cannot figure out what is wrong. – mgus Mar 13 '18 at 00:25
  • There is something wrong in your set up or calculation indeed from your results $Ap_1\neq p_1-p_2$. – user Mar 13 '18 at 00:30
  • @gimusi But can you check whether the 5 equations I setup to solve for $p_1,\dots,p_5$ are correct? If I understand correctly, I need to find $p_1,\dots,p_5$ that satisfy these equations but are also linearly independent. Is this true? – mgus Mar 13 '18 at 00:46
  • You set up seems ok and equivalent to mine, maybe you went wrong solving the systems, check again gor the calculation. – user Mar 13 '18 at 00:50

2 Answers2

1

Note that the condition $AP=PJ$ is equivalent to

  • $Ap_1=-p_1 \to p_1$
  • $Ap_2=p_1-p_2\to p_2$
  • $Ap_3=-p_3\to p_3$
  • $Ap_4=p_4 \to p_4$
  • $Ap_5=p_4+p_5 \to p_5$

Since the set up is equivalent, from you results seems that there is something wrong in the calculation indeed $Ap_1\neq p_1-p_2$.

Notably from

  • $Ap_1=-p_1 \implies (A+I)p_1=0$
  • $Ap_2=p_1-p_2\implies (A+I)p_2=p_1$

we obtain

  • $p_1=(0,1,1,0,0)$
  • $p_2=(-1,0,0,0,0)$

from

  • $Ap_3=-p_3 \implies (A+I)p_3=0$

excluding $p_1$ we obtain

  • $p_3=(1,0,0,0,1)$

and from

  • $Ap_4=p_4 \implies (A-I)p_4=0$

  • $Ap_5=p_4+p_5 \implies (A-I)p_5=p_4$

we obtain

  • $p_4=(0,0,0,-1,0)$
  • $p_5=(0,-1,0,0,0)$
user
  • 162,563
  • So if I understand correctly, in order to find $p_1,p_2$ you jointly solve the first two systems and choose $p_1,p_2$ to be linearly independent then you choose $p_3$ so that it satisfies the 1st system and is independent of $p_1,p_2$ and finally you jointly solve the last two systems of equations to find $p_4,p_5$ so that all of $p_1,\dots,p_5$ are linearly independent. Correct? – mgus Mar 14 '18 at 02:37
  • $p_1$ and $p_2$ are not independent thus you need to find p_1 which satisfies also the system for $p_2$ while $p_3$ does not work, then we first havevto find $p_1$ and then $p_3$ – user Mar 14 '18 at 07:13
0

I got the Jordan blocks in slightly different order.

What you seem to be missing is the consistency part: in my

$$ P = \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 0 \end{array} \right) $$

we have a place where we do have $(A+I)^2 p_3 = 0,$ but we have consistency in that $p_2 =(A+I)p_3 .$ It follows automatically that $(A+I)p_2 = (A+I)^2 p_3 = 0.$

We also have $(A-I)^2 p_5 = 0,$ then $p_4 =(A-I)p_5.$ As a result $(A-I) p_4 = 0.$

$$ \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & -1 & 0 & 0 \end{array} \right) \left( \begin{array}{rrrrr} -1 & 0 & 0 & 0 & 0 \\ -1 & 1 & -2 & 0 & 1 \\ -1 & 0 & -1 & 0 & 1 \\ 0 & 1 & -1 & 1 & 0 \\ 0 & 0 & 0 & 0 & -1 \end{array} \right) \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 0 \end{array} \right) = \left( \begin{array}{rrrrr} -1 & 0 & 0 & 0 & 0 \\ 0 & -1 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 \end{array} \right) $$

To get the Jordan form in the order they report, use my columns but permuted, $p_2p_3p_1p_4p_5$ and then correct $P^{-1}.$ We can correct $P^{-1}$ by permuting the rows to 23145.

Will Jagy
  • 146,052
  • I cannot see your point about consistency and why some order of Jordan blocks is better than another. I think that my setup is correct. How did you get $P$? Can you state all equations you solved? – mgus Mar 13 '18 at 04:25