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Hello I have a lot of trouble trying to put this matrix in Jordan form. I do not really understand Jordan form theory so any help or comment would be much appreciated.

$$\begin{pmatrix} 4&1&0 \\-1&2&0\\1&1&3 \end{pmatrix}$$

I already found that the characteristic polynomial is $(x-3)^3$,

that $Ker(A−3I_3)=<(0,0,1),(−1,−1,0)>$,

and $Ker(A−3I_3)^2=C^3$

The main issue is how to find a basis for my Jordan form.

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Take a vector $v$ not in $\ker(A - 3 I)$. Compute $u=(A - 3I)v$. Complete $u$ to a basis of $\ker(A - 3 I)$ with some $w$.

A basis that works is $(u,v,w)$.

The reason this works is as follows. Given that you know that that $\ker(A - 3 I)^2$ is the full space, you know that the maximal size of Jordan block is $2$. And, since $\ker(A - 3 I)$ is not the full space you know that there is a Jordan block of size greater $1$.

Thus the only way this can happen is you have one block of size $2$ and one block of size $1$.

For the Jordan block of size $2$ you need a vector $v$ not in $\ker(A - 3 I)$ itself and its image $u=(A - 3 I)v$. Note that $u = (A- 3I)v$ means that $Av= u + 3v$. Compare this to the the second colon in the Jordan block where you just have $1$ and $3$.

So with those two you have the Jordan block of size two covered. Then you need another vector in the kernel to get another block of size one.

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  • $\begingroup$ Ok it works thank you. Does this method work in general? Could you explain a little of the theory behind it? $\endgroup$ – Jonathan Baram Apr 28 '15 at 19:48
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    $\begingroup$ It always works when the dimensions are as you have it. I will work something more detailed later; $\endgroup$ – quid Apr 28 '15 at 19:53
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    $\begingroup$ I added some explanation. $\endgroup$ – quid Apr 28 '15 at 22:28

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