Linear Algebra - Jordan form basis

oferon

Hi all,
I'm having trouble finding jordan basis for matrix A, e.g. the P matrix of: [itex]J=P^{-1}AP[/itex]
Given [itex]A = \begin{pmatrix} 4 & 1 & 1 & 1 \\ -1 & 2 & -1 & -1 \\ 6 & 1 & -1 & 1 \\ -6 & -1 & 4 & 2 \end{pmatrix}[/itex]

I found Jordan form to be: [itex]J = \begin{pmatrix} -2 & & & \\ & 3 & 1 & \\ & & 3 & \\ & & & 3 \end{pmatrix}[/itex]

Now wer'e looking for [itex]v_1, v_2, v_3, v_4[/itex] such that:

[itex] Av_1 = -2v_1 → (A+2I)v_1=0[/itex]
[itex]Av_2 = 3v_2 → (A-3I)v_2=0[/itex]
[itex]Av_3 = v_2+3v_3 → (A-3I)v_3=v_2[/itex]
[itex]Av_4 = 3v_4 → (A-3I)v_4=0 [/itex]

So now I find: [itex]v_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \\ -1 \end{pmatrix} \hspace{10mm} v_2,v_4 = \begin{pmatrix} 1 \\ 0 \\ 1 \\ -2 \end{pmatrix},\begin{pmatrix} 0 \\ 1 \\ 0 \\ -1 \end{pmatrix}[/itex]

Now I try to solve [itex] (A-3I)v_3=v_2[/itex] for each of the possible v2's I just found above, but there's no solution for any of em'...

[itex]A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ -1 & -1 & -1 & -1 \\ 6 & 1 & -4 & 1 \\ -6 & -1 & 4 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}=\begin{pmatrix} 1 \\ 0 \\ 1 \\ -2 \end{pmatrix}\hspace{5mm} OR \hspace{5mm} A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ -1 & -1 & -1 & -1 \\ 6 & 1 & -4 & 1 \\ -6 & -1 & 4 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}=\begin{pmatrix} 0 \\ 1 \\ 0 \\ -1 \end{pmatrix}[/itex]


Where am I going wrong? Thanks in advance!

I like Serena

Hi oferon!

Did you consider that the proper v₂ could be a linear combination of your current v₂ and v₄?
What if you try ##\lambda v_2 + \mu v_4## to find v₃?

oferon

If it was a linear combination of other vectors then V1-4 would not be a basis.. Am I wrong?

Plus, another student told me the method I tried was completly wrong and that the correct method is finding more vectors through
[itex] Ker (A-λI)^j[/itex] where j=2,3,... depends on how many more vectors I need for my basis.

Which of the methods should I use? Any why? I'm lost

I like Serena

You need to find a ##v_3## that satisfies ##(A-3I)v_3=λv_2+μv_4##.
When you find it, v1-v4 will form a basis.


That would work too, but it seems to me that it is a lot more work.
(Short story: that student is wrong. Your method is fine. You just did not finish it.)


If you're wondering... try both?

oferon

Hi, thanks for your kind replies.

Ok, first I try what you suggested.. I take [itex] (A-3I)v_3 = λv_2+μv_4 [/itex] I get:

[itex]\begin{pmatrix} 1 & 1 & 1 & 1 \\ -1 & -1 & -1 & -1 \\ 6 & 1 & -4 & 1 \\ -6 & -1 & 4 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}=\begin{pmatrix} λ \\ μ \\ λ \\ -2λ-μ \end{pmatrix} ----> \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 6 & 1 & -4 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}=\begin{pmatrix} λ \\ λ+μ \\ λ \\ -λ-μ \end{pmatrix}[/itex]

Now I see it must satisfy [itex] μ = -λ[/itex] so I pick [itex]λ=1, μ=-1[/itex] thus [itex] v_2-v_4=\begin{pmatrix} 1 \\ -1 \\ 1 \\ -1 \end{pmatrix}[/itex] so now I solve: [itex]\begin{pmatrix} 1 & 1 & 1 & 1 \\ -1 & -1 & -1 & -1 \\ 6 & 1 & -4 & 1 \\ -6 & -1 & 4 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}=\begin{pmatrix} 1 \\ -1 \\ 1 \\ -1 \end{pmatrix}[/itex]
But the solutions I get are exactly [itex] \begin{pmatrix} 1 \\ 0 \\ 1 \\ -2 \end{pmatrix} , \begin{pmatrix} 0 \\ 1 \\ 0 \\ -1 \end{pmatrix} [/itex] The same v2,v4... So where am I wrong now?


Second thing, I've searched all over the net, and found this method. Yet the method the other student told me is what was taught in class. Can I be 100% sure both methods are equivalent and can be used both in all cases?
I thank you again for your time.

oferon

Ok, so I asked our instructor about the second question and yes, both methods are good.
I prefer "my" method, but as you can see I still get stucked with it.. So how do I move on with this [itex] (A-3I)v_3 = λv_2+μv_4 [/itex] ?
Thanks again

I like Serena

Can you find a 3rd solution that is independent of v2 and v4?
(Let's say with the first 2 entries set to zero. ;)

oferon

Hmm, ok I see what you say..
So now I have 3 final questions to close this case for good:

1) I thought all solutions were given by span of [itex] \begin{pmatrix} 1 \\ 0 \\ 1 \\ -2 \end{pmatrix} , \begin{pmatrix} 0 \\ 1 \\ 0 \\ -1 \end{pmatrix} [/itex]
So where did this [itex] \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}[/itex] (tho I agree it IS a solution for this system) come from??

2) How is it possible that [itex]v_2 , v_4[/itex] are solutions of both homogeneous and non-homogeneous
[itex](A−3I)v_3=0[/itex] and [itex](A−3I)v_3=v_2-v_4[/itex]. I doubt if it was just by accident..

3) Final question is how come I'm allowed to go from the equation I got by comparing columns of PJ and AP: [itex](A−3I)v_3=v_2[/itex], ,
to the equation [itex](A−3I)v_3=λv_2+μv_4[/itex]?
The third column in J matrix [itex] \begin{pmatrix} 0 \\ 1 \\ 3 \\ 0 \end{pmatrix}[/itex] clearly shows I should find [itex]Av_3=v_2+3v_3[/itex] , not [itex]Av_3=v_2+3v_3-v_4[/itex]

I appreciate your help alot! Thank you.

oferon

Oh ok, I discard my 3rd question... The answer is that I pick v2 to be [itex]\begin{pmatrix} 1 \\ -1 \\ 1 \\ -1 \end{pmatrix}[/itex]

Now I remain only with questions 1, and 2.. More related to equations system rather than J form I suppose

oferon

OK, please discard all of my question, I'm an idiot :)
Everything is clear now, I thank you very much for the last time :)

I like Serena

Okay... I just got around to looking at your thread again.
But it seems you've already answered your own questions.

Good!