Find a matrix $P$ such that $P^{-1}AP$ is in Jordan Normal Form.

Question

The matrix in question is:

$$A = \begin{bmatrix} 2 && -1 \\ 4 && 6 \end{bmatrix}$$

This has characteristic and minimal polynomial $p(\lambda) = (\lambda - 4)^2$

An eigenvector is $(1,-2)$, from a one dimensional eigenspace so not diagonalisable.

Don't I now have to find an eigenvector $(A-4\mathbb{I})^2v_2 = 0$?

but $(A-4\mathbb{I})^2 = 0$.

Now I am pretty stumped. I can not seem to find some good material on this.

Answer 1

You have two options.

1. Since you already have an eigenvector $v_1=(1; -2)^\top$, just find a non-zero vector $v_2$ such that $(A-4 I)v_2 = v_1$. Then $v_1,v_2$ are $1$st and $2$nd columns of matrix $P$ respectively.

2. Or, since $(A-4 I)^2$ is the zero matrix, you could take the standard impulse basis $e_1=(1,0)^\top$, $e_2=(0,1)^\top$ and try to find $w=(A-4I)e_1$. If $w \neq \vec 0$ then $w,e_1$ is your basis, otherwise $v=(A-4I)e_2$ should be non-zero and the basis is $v,e_2$.

Answer 2

The only possible Jordan normal form for a non-diagonalizable 2x2 matrix is $${\bf A} = {\bf SJS}^{-1}\hspace{1cm}\textrm{if}\hspace{1cm} {\bf J} = \left[\begin{array}{cc}\lambda & 1\\ 0&\lambda\end{array}\right]$$ So you see by inspection $${\bf P} = {\bf S}^{-1}$$ So we now search for a vector $\bf v$ is so that $${\bf Av} = \lambda {\bf v} + 1{\bf w}, \textrm{ if } {\bf w} = [1,-2]^T$$ ( The numbers 1 and $\lambda$ come from the positions in the $\bf J$ matrix above ). You can rewrite this as a linear equation system which you can solve. Then $\bf v$ and $\bf w$ will be the columns in the base change matrix.