Question

I have a question, need help
Let A in L(P_2(R) be given by Ax(t) =x'(t) . Find all ei.value and ei.vector

Answer 1

In my course, P_2 has 2 dimensions. My question is :
if A applies on the basis of P-2 ={1,t}, then
A(1) =0
A(t)=1
so that A =\(\left[\begin{matrix}0&0\\1&0\end{matrix}\right]\) so that ei.value =0 *2
eivector = (1,0)
However!!

Answer 2

if I go this way:
A: X(t) -->X'(t)
Nulity (A) = {x(t) such that X'(t)=0} ={c:c in R}
dim(Nul (A) =1--> rank (A) =1 --> 2ei.value 0 and a non-zero one.
What's wrong?

Answer 3

\(A=\left[\begin{matrix}0&1\\0&0\end{matrix}\right]\)

Answer 4

P_2 is represent a 3 dimensional space spanned by {1,t,t^2}
Do you mean P_1. That would be spanned by {1,t}

Answer 5

In my course, it is defined P2 has 2 dimension as I indicated above.

Answer 6

If you just need to find the eigenvalues of A, the matrix you wrote down, then we can do that very quickly

Answer 7

Ah okay I didn't observe that...
We can work from there. Anyway, you wrote the A matrix and said you need the eigenvalues of it, correct?

Answer 8

I want to make it clear. I was taught that to n xn matrices, we can find out n eigvalues which come from both nul(A) and range (A). The eivalues from nul(A) =0 and e.values from Rang(A) \(\neq 0\)

Answer 9

@Miracrown the problem is from the post. The part I wrote in comment box is the way I find out the solution. However, I have 2 ways give me 2 different solutions :(.

Answer 10

I'm not really following the logic in the message. Your two eigenvalues should be 0

Answer 11

You can see that obviously, the Rang(A) \(\neq 0\), so, why don't I have eigenvalue \(\neq 0\)? for example if x(t) = t, then x'(t) =1, Ax(t) =1

Answer 12

I don't see why that's a problem Why should you have an eigenvalue other than 0?

Answer 13

I am not sure what theorems you are basing this off of. Why is it obvious that the Range(A) !=0. Well if the matrix you wrote down is correct, then you should only have the repeated eigenvalue 0

Answer 14

My thought is: Av = 0 --> v \(\in nul(A)\) and eigenvalue = 0
Av\(\neq 0\) --> Av is in Rang (A) and there is some eigenvalue \(\lambda \neq 0\) such that Av =\(\lambda\)v

Answer 15

That's why there must exists some \(\lambda \neq 0\) to build up Rang (A)

Answer 16

That's not necessarily true. Just because the range is non-zero does not mean that you have non-zero eigenvalues. For example, the matrix\[\left[\begin{matrix}0&1\\0&0\end{matrix}\right]\]

Answer 17

Okay I can see now. Well, I wouldn't say that's necessarily the case though.

Answer 18

Not really. The image of A can just be some other vectors. The idea is to get Av=lv. But the image is going to be the derivative

Answer 19

So the only way we can produce that situation is either v=0, which it can't or lambda=0

Answer 20

There is no vector v(t), such that v=v', unless v=0 or lambda =0
Again we cannot have 0 for an eigenvector. So the case must be that lambda =0

Answer 21

OK, Thank you very much. Tomorrow is my final, I would like to make everything clear before it. One more question: :)
why if u'\(\in\)[u], then u'-u \(\in \) nul(A)?

Answer 22

For even further proof, you can just remember that the differential equation\[y'(t)=\lambda y(t)\]has the solution\[y=ce^{\lambda t}\]for some constant \(t\). This is not a polynomial for \(\lambda\neq0\), and so you can't have a non-zero eigenvalue.

Answer 23

*for some constant \(c\).

Answer 24

Not a prob ^^ I understand everything defined in your question except for [u]. What does [u] mean?

Answer 25

@KingGeorge You said exactly what my prof said. But I don't understand him.hishihihi

Answer 26

@Miracrown [u] is equivalent subset

Answer 27

Presumably \([u]\) is the equivalence class of \(u\) under multiplication by \(A\)?

Answer 28

yup

Answer 29

oh :O

Answer 30

If that were the case then what would be the relationship between u and u'
That is, if there was an equivalency between u and u' what would that mean

Answer 31

I presume it would be \(u\) ~ \(v\) iff \(Au=Av\).

Answer 32

Right :P !

Answer 33

So if u' is in [u] then Au'=Au
Correct?

Answer 34

yes, the conclusion is if u'-u in nul (A) then A(u'-u) in nul (A)
but i don't know why if u' in u ,then u'-u in nul(A)

Answer 35

First of all, A is a linear transformation, so it distributes across addition. So\[A(u-u')=Au-Au'\]Therefore...

Answer 36

All we need is some algebra on this equation Au'=Au. Move everything to one side...

Answer 37

ok, let me post a new one to make it clear. Thanks for being here, mathematicians. :)