# Eigen vectors and values

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

Hi PunkyC7! 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.

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)?

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...

so thats how they go (0,-1) ok
so know do you know if it consists of eigen vectors

Well, what is an eigenvector?

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

Indeed, so is T(1,2) from the form lambda*(1,2)?

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

Certain? What is T(1,2) in terms of (1,2) and (2,3)?

x=0
y=-1

So T(1,2)=-(2,3), so this indeed means that you won't have a basis of eigenvectors!

why does it mean it?

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)...

Ray Vickson