Linear Algebra proof, diagonalization

[hedgie]

Homework Statement

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))

Homework Equations

N/A

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.

 

[Office_Shredder]

I am stuck on finding the basis

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

 

[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?

 

[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?

 

[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.

 

[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?

 

[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!

 

[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?

So to answer your question, yes, it's the same basis.
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.

 

[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.

 

[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?

 

[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!

 

[I like Serena]

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

Good luck!

 

[hedgie]

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

 

[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?

 

[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...

 

[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.

 

[hedgie]

Why can we switch v1 and v2 in post 6?

 

[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"?

 

[I like Serena]

Why can we switch v1 and v2 in post 6?

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

 

[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.

 

[hedgie]

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

What I was asking here is can you swap the order of the vectors/basis. The equalities would change to the matrix you wrote early((1,0),(0,0))

 

[abr_pr90]

The matrix of a nilpotent transformation over a cyclic invariant subspace relative to the basis x, Ax, A2x, ..., Am-1x is particularly simple. It has ones just above the main diagonal and zeros elsewhere, as in this four-dimensional example:

| 0 1 0 0 |
| 0 0 1 0 |
| 0 0 0 1 |
| 0 0 0 0 |

Theorem. A nilpotent matrix is similar to a direct sum of matrices, each of which has ones just above the main diagonal and zeros elsewhere.

http://www.efgh.com/math/algebra/eigenvalue.htm

 

[hedgie]

The matrix of a nilpotent transformation over a cyclic invariant subspace relative to the basis x, Ax, A2x, ..., Am-1x is particularly simple. It has ones just above the main diagonal and zeros elsewhere, as in this four-dimensional example:

| 0 1 0 0 |
| 0 0 1 0 |
| 0 0 0 1 |
| 0 0 0 0 |

Theorem. A nilpotent matrix is similar to a direct sum of matrices, each of which has ones just above the main diagonal and zeros elsewhere.

http://www.efgh.com/math/algebra/eigenvalue.htm

We cannot use any of that. Thanks though.

 

[I like Serena]

Did you understand what I meant when I wrote: N*[ V_2 V_1 ]?
What is [ V_2 V_1 ] in this context?

 

[I like Serena]

What I was asking here is can you swap the order of the vectors/basis. The equalities would change to the matrix you wrote early((1,0),(0,0))

That is correct.

 

[hedgie]

Did you understand what I meant when I wrote: N*[ V_2 V_1 ]?
What is [ V_2 V_1 ] in this context?

[V_2 V_1] is the basis for R^2 such that N*V_1=V_2 and N*V_2=0 is true

 

[I like Serena]

[V_2 V_1] is the basis for R^2 such that N*V_1=V_2 and N*V_2=0 is true

Yes! This is true.
Did you note that [V_2 V_1] is also a matrix?

Let's define P=[V_2 V_1].
Is P invertible?

 

[hedgie]

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

So do I just guess to find the basis...say N*(0,1) then you'd find V2=(0,0)...but why does just finding a basis for R^2 show that its similar?

 

[hedgie]

Yes! This is true.
Did you note that [V_2 V_1] is also a matrix?

Let's define P=[V_2 V_1].
Is P invertible?

Since [V_2 V_1] is a matrix that would make [(0,1),(1,0] the change of coordinate matrix but why can we switch the order of V_1 and V_2.

I think it is invertible.

 

[I like Serena]

Since [V_2 V_1] is a matrix that would make [(0,1),(1,0] the change of coordinate matrix but why can we switch the order of V_1 and V_2.

I think it is invertible.

Yes, it is invertible, because V_1 and V_2 have to be linearly independent.
(If you're up to it... why?)

So let's define J=[(0,0) (1,0)].
Then we'd have N P = P J with P an invertible matrix.
Can you rewrite this?

 

[hedgie]

Yes, it is invertible, because V_1 and V_2 have to be linearly independent.
(If you're up to it... why?)

So let's define J=[(0,0) (1,0)].
Then we'd have N P = P J with P an invertible matrix.
Can you rewrite this?

So we don't know N and when I write that I get ((0,0),(0,1)) on the RHS and N*(V_2,V_1) on the LHS.

 

[I like Serena]

Yes.
What happens if you multiply the LHS and the RHS with the inverse of P?

 

[hedgie]

Yes.
What happens if you multiply the LHS and the RHS with the inverse of P?

So I get N*((1,0),(0,1))=((0,0),(1,0)) which is what we want. Because N times the identity equals the similar matrix.

So we can just use anything that generates R^2...

Isn't it a lot simpler to say N*(0,1)=(0,0) where (0,1) is V_1 as I noted in my original attempt at the problem or cannot you not do that because we do not know what N is?

 

[I like Serena]

So I get N*((1,0),(0,1))=((0,0),(0,1)) which is what we want.

Umm... no...
You need to show that N is similar to J.
But you don't and can't know what N, P, V_1, or V_2 are...
You can only find some properties of them, like to which matrix N is similar to, and that V_1 and V_2 are linearly independent.

So we can just use anything that generates R^2...

Isn't it a lot simpler to say N*(0,1)=(0,0) where (0,1) is V_1 as I noted in my original attempt at the problem or cannot you not do that because we do not know what N is?

N*(0,1)=(0,0) may or may not be true. It's not possible to say.

 

[hedgie]

Why do we not know V_1 and V_2 are they not E_1 and E_2 for R^2 and then transformed to get ((0,1),(0,0))