# Finding a jordan basis for a jordan form

Let $$A = \left(\begin{array}{cccc} 1&0&0&0\\3&-2&0&0\\14&0&-2&0\\8&-1&1&-2 \end{array}\right)$$

Easy to verify that $f_(x) = (x-1)(x+2)^3$. So the eigen-values are: $\lambda = 1,-2$. Some more calculations gives the result that the Jordan form is:

$$J = \left(\begin{array}{cccc} 1&0&0&0\\0&-2&0&0\\0&0&-2&0\\0&0&1&-2 \end{array}\right)$$

Now, I want to find $P$ such that $P^{-1}AP = J$. Meaning, we wish to find $AP = PJ$. More detailed. Let's denote $P=(v_1,v_2,v_3,v_4)$. Then,

$$Av_1 = v_1 \implies (A-I)v_1 = 0 \\ Av_2 = -2v_2 \implies (A+2I)v_2 = 0 \\ Av_3 = -2v_3 + v_4 \implies (A+2I)v_3 = v_4 \\ Av_4 = -2v_4 \implies (A+2I)v_4 = 0$$

Now, that means that $v_3 \in \ker (A+2I)^2$ since $$(A+2I)^3v_3 = (A+2I)(A+2I)v_3 = (A+2I)v_4 = 0$$

I evaluated that:
$$\ker(A+2I) = \text{span}\{(0,0,0,1), (0,1,1,0)\}$$ $$\ker(A+2I)^2 = \text{span}\{(0,0,0,1), (0,1,0,0), (0,0,1,0)\}$$

First we choose $v_4\in\ker(A+2I)$. We may select $v_4 = (0,0,0,1)$.

Now, we want $v_3$ such that $$(A+2I)v_3 = v_4$$

So I picked $v_3 = (0,1,1,0)$ which answers the criteria.

But now, I need to pick $v_2,v_1$ which linearly independent of $v_3,v_4$ but that's impossible I'm afraid.

What am I doing wrong/ understand wrong?

• "some more calculations gives the result that the Jordan form is..." usually we determine jordan form and basis together – Michael Galuza Jul 22 '15 at 15:41
• @MichaelGaluza dim(Ker($A+2I$))=$2$ gives $2$ Jordan blocks with eigenvalue $-2$ without calculation of the basis. – A.Γ. Jul 22 '15 at 15:51
• @jmiller You cannot pick $v_3=(0,1,1,0)$ since it belongs to $\text{Ker}\,(A+2I)$ and gives $v_4=0$. $v_3$ should be in $\text{Ker}\,(A+2I)^2\setminus\text{Ker}\,(A+2I)$. – A.Γ. Jul 22 '15 at 16:01