Linear Algebra proof, diagonalization

1. Nov 29, 2011

hedgie

1. The problem statement, all variables and given/known data

Let N be a 2x2 matrix such that N^2 = 0. Prove that either N = 0 or N is similar to the Matrix
((0,0),(1,0))

2. Relevant equations

N/A

3. The attempt at a solution

N^2=0
Assume N ≠0
Show N is similar to ((0,0),(1,0))

Need to find a basis of R^2 {V_1,V_2}

where:
NV_1=V_2
NV_1=0

Then N^2(V_1)=N(V_2)=0

I am stuck on finding the basis would I just use N*(1,1)=(0,1) and N*(0,1)=(0,0) giving me the same matrix..........or do I need to find a similar matrix using the definition =Q^-1*A*Q

Thanks in advanace and hope I am not being to unclear.

2. Nov 29, 2011

Office_Shredder

Staff Emeritus
As you said, we have N(v1)=v2, so all you really need to do is find v1. The requirements for what this vector needs to satisfy are pretty simple... remember, at some point you need to use the fact that N=/=0

3. Nov 29, 2011

hedgie

Can I just use the basis (1,1) and (0,1)

Then I'd get N*(1,1)=(0,1) and N*(0,1)=(0,0)
giving me the same matrix..........or do I need to find a similar matrix using the definition D=Q^-1*N*Q the matrix?

4. Nov 29, 2011

I like Serena

What can you say about the eigenvalues of N^2?
And therefore of the eigenvalues of N?

Do you know what a Jordan normal form is?

5. Nov 29, 2011

hedgie

I do not know what Jordan normal form is yet....we have not learned that.

I don't know what I could say about the eigenvalues. Obviously in the matrix given they are 0. But they could be +-1 in a similar matrix. Which is why I am lost I am unable to solve the simple N*V_1=V_2 and N*V_2=0. I get how it is in the original equation. But am brain dead or something on solving it.

6. Nov 29, 2011

I like Serena

If N*V_1=V_2 and N*V_2=0, then
V_1 and V_2 are linearly independent, and so [ V_2 V_1 ] is invertible.

Furthermore, we have:

N*[ V_2 V_1 ]=[ N*V_2 N*V_1 ] = [ 0 V_2 ] = [ V_1 V_2 ]*[ (0,0) (1,0) ]

Can you draw a conclusion from this?

7. Nov 30, 2011

hedgie

I have stared at this for awhile.

I feel that i understand it but still do not seed the basis. I see that you swapped the order of the linearly independent set but that is the same basis correct??

I appologize I know this is a really simple problem and I appreciate your help!

8. Nov 30, 2011

I like Serena

Yes, I knew it was a bit of a drop on you and that there is perhaps a bit much condensed information in there.
My problem is that I do not know exactly what you already know and understand and what not.
Can you enlighten me?

I swapped the vectors because otherwise I'd get [ (1,0) (0,0) ] instead of [ (0,0) (1,0) ], and your problem asks for the latter.

9. Nov 30, 2011

hedgie

OK, good then I was able to follow everything you wrote in the prior post. It was condensed because it was helpful to think through it.

So are you asking me what I have learned/retained? or what we have covered in our course. In our course we are using Friedberg and have finished through 5.2 excluding the *'d sections, so diagonalization and diagonailzability. So we are doing eigen vectors and characteristic polynomials now. Should I be able to see something with the eigenvalue being 0 for the matrix N and its similar matrix? Or is there something else that I should understand that I missed?

Thanks so much for all your help.

10. Nov 30, 2011

I like Serena

You write: "I don't know what I could say about the eigenvalues. Obviously in the matrix given they are 0. But they could be +-1 in a similar matrix."

Which "matrix given" do you mean?
And why do you think they could be +-1 in a similar matrix?

11. Nov 30, 2011

hedgie

The same matrix but with -1 instead of 1. I should have mentioned we did do the review of Complex variables a week or two ago.....Also any advice for how to proceed after this course to learn more from Friedberg, because there is not another course offered at our school. And we did not make it as far as I hoped in the book! I guess just slog through it and ask problems when I'm stuck.

Thanks!

12. Nov 30, 2011

I like Serena

Anyway, the eigenvalues can not be +-1. They will always be zero.

Good luck!

13. Nov 30, 2011

hedgie

Sorry. Didn't realize I wrote Eigen values. Thanks!

Last edited: Nov 30, 2011
14. Nov 30, 2011

hedgie

Ok sorry I am still confused.

My problem is the matrix ((0,1),(0,0) where those are columns. I know that q^-1*A*q=B and q is supposed to be ((0,1),(1,0)) and B is supposed to be ((0,0),(1,0)).

How do I solve to find q and though? And how do I know that is B?

15. Nov 30, 2011

I like Serena

q is not supposed to be what you write.
And which matrix is B?
Haven't seen it before.

Btw, in post #6 we were not done yet...

16. Nov 30, 2011

hedgie

I am just using the formula to find a similar matrix which is q^-1*A*q=B.

A is the matrix I am given the problem and B would be N a similar matrix.

17. Nov 30, 2011

hedgie

Why can we switch v1 and v2 in post 6?

18. Nov 30, 2011

I like Serena

Huh?
There is no matrix A given in the problem.
And what do you mean with "B would be N a similar matrix"?
Did you perhaps intend: "B would be a similar matrix to N"?

19. Nov 30, 2011

I like Serena

Of course, but then the equalities have to change as well.
Can you say what the equalities would become?

20. Nov 30, 2011

hedgie

OK, there is no matrix A. I am labeling the matrix given in the problem as A and then using the formula for a similar matrix which in my book is B=Q^-1*A*Q.....I do not know what Q or B is. I know what they are supposed to be by guessing but how does one solve for Q and B.