What is wrong with my calculation of Jordan canonical form?

Question

I want to calculate the Jordan canonical form of $L_A=\begin{pmatrix}2&2&1\cr0&3&0\cr1&-1&2\end{pmatrix}$

First find the eigenvalues $\lambda_1=1,\lambda_2=3$

$N(L_A-\lambda_1 I)=\{x_3(1,0,1)\}$

$L_A-\lambda_2 I=\begin{pmatrix}-1&2&1\cr0&0&0\cr1&-1&-1\end{pmatrix}$

$(L_A-\lambda_2 I)^2=\begin{pmatrix}2&-3&-2\cr0&0&0\cr-2&3&2\end{pmatrix}$

I want to find $v$ such that $(L_A-\lambda_2 I)^2v=0$ (i) and $(L_A-\lambda_2 I)v\ne0$ (ii) since $\lambda_2$ can only have one cycle in a Jordan basis (otherwise $\lambda_2$ would have two eigenvalues, a contradiction to the dimension of its eigenspace)

Let $v=(x_1,x_2,x_3)$ then by (i)

$$x_1=\frac{3}{2}x_2+x_3,x_2=x_2,x_3=x_3$$

We also consider (ii). $L_A-\lambda_2 I$ times such vector $\ne0$ gives

$$(\frac{1}{2}x_2+2x_3,0,-\frac{1}{2}x_2)\ne0$$

One example is to choose $x_2=2,x_3=0$. So $v=(3,2,0)$, $(L_A-\lambda_2I)v=(1,0,-1)$. A Jordan basis is $\gamma=\{(1,0,1),(1,0,-1),(3,2,0)\}$. Now find the Jordan form w.r.t. this basis.

$$[L_A(1,0,1)]_\gamma=(3,0,0)$$

$$[L_A(1,0,-1)]_\gamma=(0,1,0)$$

$$[L_A(3,2,0)]_\gamma=(1,0,3)$$

This gives

$$\begin{pmatrix}3&0&1\cr0&1&0\cr0&0&3\end{pmatrix}$$

But this is not a Jordan form, why?

Answer 1

HINT

Jordan normal form is as follow

$$J=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 3 \\ \end{pmatrix}$$

By Jordan theorem we know that a matrix $P$ exists such that $$P^{-1}L_AP=J$$

let $$P=[v_1,v_2,v_3]$$

then P has to satisfy the following system: $$L_AP=PJ$$ that is in this case $$L_Av_1=v_1\implies (L_A-I)v_1=0$$ $$L_Av_2=3v_2\implies (L_A-3I)v_2=0$$ $$L_Av_3=v_2+3v_3\implies (L_A-3I)v_3=v_2$$

Once we have $v_1$ we can find $v_2$ and finally $v_3$, that is

$$(L_A-I)v_1=0 \implies \begin{pmatrix}1&2&1\cr0&2&0\cr1&-1&1\end{pmatrix}v_1=0 \implies v_1=(1,0,-1)$$

$$(L_A-3I)v_2=0 \implies \begin{pmatrix}-1&2&1\cr0&0&0\cr1&-1&-1\end{pmatrix}v_2=0 \implies v_2=(1,0,1)$$

$$(L_A-3I)v_3=v_2 \implies \begin{pmatrix}-1&2&1\cr0&0&0\cr1&-1&-1\end{pmatrix}v_3=v_2 \implies v_3=(3,2,0)$$

and thus

$$P=\begin{pmatrix} 1 & 1 & 3 \\ 0 & 0 & 2 \\ -1 & 1 & 0 \\ \end{pmatrix}$$

Answer 2

You did find a Jordan basis for $L_A$, but then something went wrong in your final calculations. You haven’t shown any details of that, so I’m not going to try to guess what the error might have been. If you compute $$\begin{pmatrix}-1&1&3\\0&0&2\\1&1&0\end{pmatrix}^{-1} \begin{pmatrix}2&2&1\\0&3&0\\1&-1&2\end{pmatrix} \begin{pmatrix}-1&1&3\\0&0&2\\1&1&0\end{pmatrix} = \begin{pmatrix}1&0&0\\0&3&1\\0&0&3\end{pmatrix}$$ you get the expected Jordan matrix.