# Checking why I got this Jordan Canonical Form incorrect

I've redone this a few times and I still get an incorrect answer (every matrix multiplication is verified via software, so I don't think its a computation error).

For:
$$A = \begin{pmatrix} 4 & -2 & -1 \\ -2 & 3 & 1\\6 & -5 & -1\end{pmatrix}$$

I know the Jordan Block for this is:

$$J = \begin{pmatrix} 2 & 1 & 0\\ 0 & 2 & 1\\ 0 & 0 & 2\end{pmatrix}.$$

This tells me that for the matrix $P = \left[ v_1 \ \ v_2 \ \ v_3\right]$ such that $$A = PJP^{-1},$$

$$Av_1 = 2v_1 \\ Av_2 = v_1 + 2v_2 \\ Av_3 = v_2 + 2v_3.$$

Hence, for $v_1$, I just pick a vector in the kernel of $A-2I$. This will be $\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$.

Then for $v_2$, this is a vector in the kernel of $(A-2I)^2$ but not in kernel of $A-2I$. This is $\begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}$.

Then for $v_3$, this is a vector in the kernel of $(A-2I)^3 = \mathbb{C}^{3}$ that's not in the kernel of the previous two powers of $A-2I$.
I pick $v_3 = \begin{pmatrix} 0 \\ 0 \\ 1\end{pmatrix}$.

However, when multiplying out the matrix $PJP^{-1}$, it doesn't give me $A$.

## Computations

The row-reduced-echelon forms for each power of $A-2I$ are

$$A-2I \sim \begin{pmatrix} 1 & 0 & -0.5\\0 & 1 & 0\\0 & 0 & 0\end{pmatrix} \\ (A-2I)^2 \sim \begin{pmatrix}1 & -0.5 & -0.5 \\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix} \\ (A-3I)^3 \sim \begin{pmatrix} 0 & 0 & 0 \\0 & 0 & 0\\0 & 0 & 0\end{pmatrix}$$

Is my logic wrong in finding $P$?

Your equations say you want $(A-2I)v_2=v_1$, you have to choose $v_2$ that satisfies this, your choice doesn't.
• I see. That's weird, the method that I use (it's probably similar) is that I just pick basis vectors of the kernel of the powers of $A-2I$ (in this case, the power is $2$), if you're familiar with it. Do you know why it doesn't work here (simply picking vectors in kernel of $(A-2I)^2$ but not in kernel of $A-2I$)? – Natash1 Jun 14 '17 at 8:48
• The vectors from which you are choosing form a one-dimensional vector space, that is, they are all scalar multiples of (in this case) $(1,2,0)$. But only one of those multiples staisfies $(A-2I)v_2=v_1$, which is what you need in order to get $PJP^{-1}=A$. – Gerry Myerson Jun 14 '17 at 8:52
• Ah I see, so I guess I do need to do that multiplying method (firstly picking a vector, then just sequentially multiplying by $A-2I$). I could've sworn "blindly" picking vectors in say $(A-3I)^3$ but not in the lower powers gives one of the vectors of $P$, but I guess my memory is off. Thanks! – Natash1 Jun 14 '17 at 9:06