Linear Algebra - Jordan form basis

[oferon]

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

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

Now wer'e looking for v_1, v_2, v_3, v_4 such that:

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

So now I find: 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}

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

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}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 completely wrong and that the correct method is finding more vectors through
Ker (A-λI)^j 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]

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

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.

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

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

If you're wondering... try both?

 

[oferon]

Hi, thanks for your kind replies.

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

\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}

Now I see it must satisfy μ = -λ so I pick λ=1, μ=-1 thus v_2-v_4=\begin{pmatrix} 1 \\ -1 \\ 1 \\ -1 \end{pmatrix} so now I solve: \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}
But the solutions I get are exactly \begin{pmatrix} 1 \\ 0 \\ 1 \\ -2 \end{pmatrix} , \begin{pmatrix} 0 \\ 1 \\ 0 \\ -1 \end{pmatrix} 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 (A-3I)v_3 = λv_2+μv_4 ?
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 \begin{pmatrix} 1 \\ 0 \\ 1 \\ -2 \end{pmatrix} , \begin{pmatrix} 0 \\ 1 \\ 0 \\ -1 \end{pmatrix}
So where did this \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} (tho I agree it IS a solution for this system) come from??

2) How is it possible that v_2 , v_4 are solutions of both homogeneous and non-homogeneous
(A−3I)v_3=0 and (A−3I)v_3=v_2-v_4. 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: (A−3I)v_3=v_2, ,
to the equation (A−3I)v_3=λv_2+μv_4?
The third column in J matrix \begin{pmatrix} 0 \\ 1 \\ 3 \\ 0 \end{pmatrix} clearly shows I should find Av_3=v_2+3v_3 , not Av_3=v_2+3v_3-v_4

I appreciate your help alot! Thank you.

 

[oferon]

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

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!