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Linear Algebra  Jordan form basis 
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#1
Jun2912, 05:17 PM

P: 30

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 → (A3I)v_2=0[/itex] [itex]Av_3 = v_2+3v_3 → (A3I)v_3=v_2[/itex] [itex]Av_4 = 3v_4 → (A3I)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] (A3I)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! 


#2
Jun2912, 06:07 PM

HW Helper
P: 6,189

Hi oferon!
Did you consider that the proper v_{2} could be a linear combination of your current v_{2} and v_{4}? What if you try ##\lambda v_2 + \mu v_4## to find v_{3}? 


#3
Jul1012, 10:37 PM

P: 30

If it was a linear combination of other vectors then V14 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 (AλI)^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 


#4
Jul1112, 02:43 PM

HW Helper
P: 6,189

Linear Algebra  Jordan form basis
When you find it, v1v4 will form a basis. (Short story: that student is wrong. Your method is fine. You just did not finish it.) 


#5
Jul1212, 12:15 AM

P: 30

Hi, thanks for your kind replies.
Ok, first I try what you suggested.. I take [itex] (A3I)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 \\ 2λμ \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 \\ λμ \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_2v_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. 


#6
Jul1212, 12:05 PM

P: 30

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] (A3I)v_3 = λv_2+μv_4 [/itex] ? Thanks again 


#7
Jul1212, 12:25 PM

HW Helper
P: 6,189

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


#8
Jul1312, 10:20 PM

P: 30

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 nonhomogeneous [itex](A−3I)v_3=0[/itex] and [itex](A−3I)v_3=v_2v_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_3v_4[/itex] I appreciate your help alot! Thank you. 


#9
Jul1312, 11:47 PM

P: 30

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 


#10
Jul1412, 01:06 AM

P: 30

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


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