Thread: matrix representation of the linear transformation

Hello,

I need some help in understanding this problem...please someone make me understand this!

Let T: R^2---> R^2 be defined by T ((a1, a2)) = (a1 + a2, a1 - a2).

Let Beta = {(1,0), (0,1)} be an ordered basis..

A) Determine [T] Beta, the matrix representation of this linear transformation.
B) Explain in brief why T is an Isomorphism.
C) Determine [T inverse] Beta.
D) Use [T inverse] B to computer T inverse ((b1, b2)).

Originally Posted by vedicmath

Hello,

I need some help in understanding this problem...please someone make me understand this!

Let T: R^2---> R^2 be defined by T ((a1, a2)) = (a1 + a2, a1 - a2).

Let Beta = {(1,0), (0,1)} be an ordered basis..

A) Determine [T] Beta, the matrix representation of this linear transformation.
B) Explain in brief why T is an Isomorphism.
C) Determine [T inverse] Beta.
D) Use [T inverse] B to computer T inverse ((b1, b2)).

What have you tried? Recall that to find you write and so that by definition . Now, what methodology did I use to get that? Why is it now obvious that is invertible?

note that if we write e1 = (1,0), e2 = (0,1) then (a1,a2) = (a1,0) + (0,a2) = a1(1,0) + a2(0,1) = a1e1 + a2e2.

so what can we say about T(a1,a2) by the linearity of T? if A is a matrix for T (that is [T]β), what would A(e1) and A(e2) have to be (rows? columns? think about this)?

T is an isomorphism iff it is a linear bijection. well, matrices are linear, almost by default, as a result of how matrix multiplication is defined.

but what does it mean for a matrix to be injective (think about null(A), is this a subspace of R^2?)? what does it mean for a matrix to be surjective, does this have anything

to do with rank(A)? can you determine null(A) and rank(A) by inspection (you should be able to)?

finally, if T is an isomorphism, it has to have an inverse. does this mean that [T]β is invertible, as a matrix? how might you calculate it?