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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?

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  • $\begingroup$ Are you sure that $(1,0,1)$ -- I assume you mean $(1,0,1)^T$ -- is a $1$-eigenvector? I get eigenvalue $3$. And in the next bit I get $(1,0,-1)^T$ ... $\endgroup$ – ancientmathematician Apr 19 '18 at 12:28
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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.

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  • $\begingroup$ I added some details. While matrix multiplication is one way of obtaining the matrix representation w.r.t. another basis, the matrix can also be obtained column by column by considering each basis being transformed to. Eventually my result is similar to yours, it has two threes and two ones. But it's just not a Jordan form. Can you please explain? $\endgroup$ – ZHU Apr 20 '18 at 10:25
  • $\begingroup$ @ZHU Your column-by-column calculation is essentially matrix multiplication. Quite plainly, you’re making a simple computational error somewhere. The image of $(3,2,0)^T$ is $(10,6,1)^T$, which in the Jordan basis is $(0,1,3)^T$ as required. As I said before, there’s n detail about that calculation and I’m not going try to guess how you went wrong. $\endgroup$ – amd Apr 20 '18 at 10:59
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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

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

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  • $\begingroup$ Can you point out specifically what's wrong with my calculation? $\endgroup$ – ZHU Apr 19 '18 at 9:45
  • $\begingroup$ I'm sorry but I can't follow your method to derive Jordan basis, then I suggested the method I usually follow. $\endgroup$ – gimusi Apr 19 '18 at 9:49
  • $\begingroup$ That make sense. But I followed an example of FIS linear algebra book, and I want to know where am I wrong. $\endgroup$ – ZHU Apr 19 '18 at 10:06
  • $\begingroup$ Does $v_1,v_2,v_3$ has to be linearly independent? Do we need to enforce this? $\endgroup$ – ZHU Apr 19 '18 at 10:42
  • $\begingroup$ @ZHU yes of course and the existence of such basis is guaranteed by Jordan theorem $\endgroup$ – gimusi Apr 19 '18 at 12:09

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