Eigen vectors and values

  • Thread starter Punkyc7
  • Start date
  • #1
420
0
For each of the following linear operators T on a vector space V and ordered bases beta, compute [T]beta, and determine whether beta is a basis conisting of eigen vectors of T.

V=R^2, T((a,b)^t)= (10a-6b
17a-10b)

and beta ={(1,2)^t , (2,3)^t)


im using transpose because im not sure how to make thenbe vectors going down


My question is how do you do it. My back doesnt give any example on how to these types of problems. I thought of using e1 and e2 but that doesnt get me the answer in the back of the book and i tried plugging in the basis but that didnt work.


If it helps the answer for [T]beta =(02
-10)
but i have no idea how they got that
 

Answers and Replies

  • #2
22,089
3,286
Hi PunkyC7! :smile:

What is [T]beta? Is it the matrix of T with respect to beta?

If so, then denote beta={x,y}. You will have to calculate T(x) and T(y) and write these things as a linear combination of x and y. Then you need to put it into a matrix.

If it isn't clear, I'll give an example.
 
  • #3
420
0
yeah its with respect to beta

so if you plug in our first beta we get (-2,-3)^t do we set that equal to x(1,2)+y(2,3)?
 
  • #4
22,089
3,286
yeah its with respect to beta

so if you plug in our first beta we get (-2,-3)^t do we set that equal to x(1,2)+y(2,3)?
Well, you want to find x and y such that (2,3)=(x+2y,2x+3y). This is a system of two equations and two unknowns...
 
  • #5
420
0
so thats how they go (0,-1) ok
so know do you know if it consists of eigen vectors
 
  • #6
22,089
3,286
Well, what is an eigenvector?
 
  • #7
420
0
an eigen vector is a vector is a vector in V that is non zero such that T(v)=lamdav where lamda is the eigen value
 
  • #8
22,089
3,286
Indeed, so is T(1,2) from the form lambda*(1,2)?
 
  • #9
420
0
no i guess not so, since we dont have any other vector to make it span R^2 there is know reason to check the other right
 
  • #10
22,089
3,286
Certain? What is T(1,2) in terms of (1,2) and (2,3)?
 
  • #11
420
0
x=0
y=-1
 
  • #12
22,089
3,286
So T(1,2)=-(2,3), so this indeed means that you won't have a basis of eigenvectors!
 
  • #13
420
0
why does it mean it?
 
  • #14
22,089
3,286
Because T(1,2) is not a multiple from (1,2). That is T(1,2)=-(2,3), and thus not T(1,2)=lambda*(1,2)...
 
  • #15
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
You can express e1 = <1,0> and e2 = <0,1> in terms of beta[1] and beta[2]. Then anything of the form a*e1 + b*e2 can be re-written as a linear combination of beta[1] and beta[2].

RGV
 

Related Threads on Eigen vectors and values

  • Last Post
Replies
10
Views
1K
Replies
4
Views
9K
Replies
6
Views
1K
  • Last Post
Replies
2
Views
1K
Replies
3
Views
3K
  • Last Post
Replies
14
Views
4K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
5
Views
1K
Top