this is my first question here.
I'm having trouble understanding Jordan canonical form, specifically the order of vectors in the change of basis matrix, I read another question here that didn't make it clear for me and I tried experimenting with this to see if I could understand:
$A=\begin{pmatrix}2&0&0&0\\0&2&0&0\\0&0&2&1\\1&0&0&2\end{pmatrix}$
So, here we have:
$null(A-2I) = span\{\begin{bmatrix}0\\1\\0\\0\end{bmatrix};\begin{bmatrix}0\\0\\1\\0\end{bmatrix}\}$
$null((A-2I)^2) = span\{\begin{bmatrix}0\\1\\0\\0\end{bmatrix};\begin{bmatrix}0\\0\\1\\0\end{bmatrix};\begin{bmatrix}0\\0\\0\\1\end{bmatrix}\}$
$null((A-2I)^3) = span\{\begin{bmatrix}1\\0\\0\\0\end{bmatrix};\begin{bmatrix}0\\1\\0\\0\end{bmatrix};\begin{bmatrix}0\\0\\1\\0\end{bmatrix};\begin{bmatrix}0\\0\\0\\1\end{bmatrix}\}$
In that question, they used the Jordan Matrix
$J=\begin{pmatrix}2&1&0&0\\0&2&1&0\\0&0&2&0\\0&0&0&2\end{pmatrix}$
But I wanted to try and change it a bit so I used $J=\begin{pmatrix}2&0&0&0\\0&2&1&0\\0&0&2&1\\0&0&0&2\end{pmatrix}$
(I swapped the boxes)
And I ordered the vectors of the base differently to make it work
$B=\{(0,1,0,0);(0,0,1,0);(0,0,0,1);(1,0,0,0)\}$
So the change of basis matrix now is
$P=\begin{pmatrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\1&0&0&0\end{pmatrix}$
I thought that $A=PJP^{-1}$
. But instead, $PJP^{-1}=$$\begin{pmatrix}2&1&0&0\\0&2&1&0\\0&0&2&0\\0&0&0&2\end{pmatrix}$, which is the Jordan Matrix with the boxes in a different order, the one they used in the other question, and I also noticed that if instead of doing $PJP^{-1}$ I did $P^{-1}JP$, I actually got A, why does this happen?
That's what I'm trying to understand and I'm looking for a clear way to understand the way to proceed when trying to get the change of basis matrix $P$ such that $A=PJP^{-1}$, especially the order of the vectors depending on the Jordan Form I choose
Best Answer
If you want to get $$J=\begin{pmatrix}2&0&0&0\\0&2&1&0\\0&0&2&1\\0&0&0&2\end{pmatrix}$$ your matrix $P$ must be $$P=\begin{pmatrix}0&0&0&1\\1&0&0&0\\0&1&0&0\\0&0&1&0\end{pmatrix}$$
The idea is that if $ P= \begin{pmatrix} p_1 & \ p_2 & \ p_3 & p_4 \end{pmatrix}$ where $p_1,p_2,p_3,p_4$ are columns, then you want \begin{align*} (A-2I)p_4=p_3 \\ (A-2I)p_3=p_2 \\ (A-2I)p_2=0 \end{align*}
This is because you want $$A\begin{pmatrix} p_1 & \ p_2 & \ p_3 & p_4 \end{pmatrix}$ = AP = PJ = \begin{pmatrix} 2p_1 & \ 2p_2 & \ p_2+2p_3 & p_3+2p_4 \end{pmatrix}$$