# Finding Jordan-Canonical-Form of a matrix

just started Jordan forms and not sure what the general method is. I watched MathdoctorRob on YouTube, but can't really make of a method to find JCF.
Would really appreciate someone to check why my working doesn't work, and how I would have done it.

$A = \begin{pmatrix} 10 & 1 \\ -9 & 4 \end{pmatrix}$
I found the characteristic polynomial (and also the minimal polynomial) to be $m_A(\lambda) = p_A(\lambda) = (\lambda - 7)^2$, which indicates the the Jordan Block of the matrix is
$J = \begin{pmatrix} 2 & 1 \\ 0 & 2\end{pmatrix}$.

If I understand correctly, for the matrix $P$ with columns $v_1,v_2$, then the block is telling me that
$Av_1 = 7v_1$ and $Av_2 = v_1 + 7v_2$.
Now I know I can take the second equality to solve for $v_2$ and we're done, but I want to see why this doesn't work:
Since $ker(A-7I) \subseteq ker((A-7I)^2)$, then I want a vector $v_2$ such that $(A-7I)^2v_2 = 0$ and $(A-7I)v_1 \neq 0$.
The bases for $ker(A-7I)$ and $ker((A-7I)^2)$ respectively are $span\left\{\begin{pmatrix} -1 \\ 3 \end{pmatrix}\right\}$ and $span\left\{\begin{pmatrix} 1 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 1 \end{pmatrix}\right\}$ (since $(A-7I)^2 = 0$ anyway).
Then if I pick $v_2$ to be $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$, which isn't in the kernel of $A-7I$, I have
$P = \begin{pmatrix} -1 & 1 \\ 3 & 0 \end{pmatrix}$, which when I test doing
$PJP^{-1}$, it doesn't equal to $A$.
Can someone guide me?

• You just have to find a vector $v$ such that $Av= \begin{pmatrix} -1 \\ 3 \end{pmatrix} + 2v$ (Think of these coefficients as the coordinates you need, 1 and 2) – I.Padilla May 21 '17 at 4:26
• I meant $Av= \begin{pmatrix} -1 \\ 3 \end{pmatrix} + 7v$ – I.Padilla May 21 '17 at 4:41

We need to find a vector $v$ such that $$Av= \begin{pmatrix} -1 \\ 3 \end{pmatrix} + 7v$$ Note that $v$ will not be in $\ker(A-7I)$ but it will be in $\ker(A-7I)^2$ since $$(A-7I)v=\begin{pmatrix} -1\\3\end{pmatrix}$$ $$(A-7I)^2v=(A-7I)\begin{pmatrix} -1\\3\end{pmatrix}=0$$ Solving the $2$x$2$ system for $v$ you get $v$=\begin{pmatrix} {-\frac{1}{3}}\\0\end{pmatrix}So your matrix P would be $$P=\begin{pmatrix} -1 &-\frac{1}{3} \\ 3 &0\end{pmatrix}$$
• Thanks. It seems that your $P$ doesn't satisfy $PJP^{-1} = A$. – Twenty-six colours May 21 '17 at 5:13