Question

T::R2->R2, T(x1,x2) =(x-2y,2y-x). a) verify that this function is linear transformation. b)find the standard matrix for this linear transformation. Determine the ker(T) and the range(T). D) is this linear combo one to one? how about onto? what else could we possibly call it?

Answer 1

Given, T : R² R² is defined by T(x,y) = (x-2y,2y-x).

a) Let, A = (x,y) R² and B = (u,v) R².

Then, T(A) = (x-2y,2y-x) and T(B) = (u-2v,2v-u)

A+B = (x+u,y+v)

T(A+B) = T(x+u,y+v) = ((x+u)-2(y+v),2(y+v)-(x+u))

i.e., T(A+B) = ((x-2y)+(u-2v),(2y-x)+(2v-u))

i.e., T(A+B) = (x-2y,2y-x)+(u-2v,2v-u)

i.e., T(A+B) = T(A) + T(B)

Let c R. Then, cA = (cx,cy)

T(cA) = ((cx)-2(cy),2(cy)-(cx))

i.e., T(cA) = (cx-2cy,2cy-cx)

i.e., T(cA) = c(x-2y,2y-x)

i.e., T(cA) = c*T(A)

Thus T(A+B) = T(A)+T(B) for all A,B R² and T(cA) = c*T(A) for all c R and A R².

Hence, T is a linear transformation.

b) Given, T(x,y) = (x-2y,2y-x)

i.e., =

i.e., =

Therefore, the standard matrix for this linear transformation is .

c) Ker(T) = {(x,y) R² : T(x,y) = (0,0)}.

Let (u,v) Ker(T).

Then, u-2v = 0 and 2v-u = 0

i.e., u = 2v

Let us take v = k. Then, u = 2k.

Therefore, (u,v) = (2k,k) = k(2,1), k R.

Hence, Ker(T) = Span{(2,1)}.

Let M be a arbitrary vector in Im(T).

Then, M = (x-2y,2y-x)

i.e., M = x(1,-1)+y(-2,2)

Thus M is a linear combination of the vectors (1,-1) and (-2,2).

Hence, Im(T) = Span{(1,-1),(-2,2)}.

d) Since Ker(T) (0,0), therefore T is not one-to-one.

Since, dimension of Im(T) dim(R²), therefore T is not onto.

Hence, T is not an isomorphism.