# Jordan Canonical form of a $2x2$ matrix, finding matrix $P$.

So, I need to reduce the matrix $$A = \begin{bmatrix} 4 & -1 \\ 9 & -2 \end{bmatrix}$$ into its Jordan Canonical form. So, I need to find matrix $P$ such that $P^{-1}AP$ is a Jordan matrix.

So I found that $\lambda = 1$ so $v_1 = \begin{bmatrix} \frac{1}{3} \\ 1 \end{bmatrix}$. But now I need to find $v_2$ for the matrix $P$. I know that $P^{-1}AP$ should be $\begin{bmatrix} \ 1 & 1 \\ 0 & 1 \end{bmatrix}$. And I know that through solving $(A-\lambda I)x=v_1$ and letting $x=v_2$, I then got:

$x_1 = \frac{1}{9}+\frac{1}{3}t$ and $x_2=t$ but I'm not sure how to get $v_2$ out of this because everything that I try doesn't lead to the Jordan matrix.. Am I doing something wrong?

• You can let $t$ be anything you like (for example: zero), and then $v_2=(x_1,x_2)$. If you've done your calculations right, this should work. If it doesn't work, show us your calculations, maybe we'll find where things went wrong. – Gerry Myerson Feb 6 '15 at 6:12
• @GerryMyerson Oh okay. Thanks :) Yeah i got the right answer now – Sofia June Feb 6 '15 at 6:26
• Avoid fractions in the answers: taking $v_1=\begin{bmatrix}1\\3\end{bmatrix}$ works just as fine. – Bernard Feb 6 '15 at 10:47