# Linear Algebra - Jordan form basis

1. Jun 29, 2012

### 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!

Last edited: Jun 29, 2012
2. Jun 29, 2012

### 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?

3. Jul 10, 2012

### 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
$Ker (A-λI)^j$ 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

Last edited: Jul 10, 2012
4. Jul 11, 2012

### 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?

5. Jul 12, 2012

### 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.

6. Jul 12, 2012

### 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

7. Jul 12, 2012

### 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. ;)

8. Jul 13, 2012

### 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.

9. Jul 13, 2012

### 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

10. Jul 14, 2012

### oferon

Everything is clear now, I thank you very much for the last time :)

11. Jul 14, 2012