Vectors: linear transformation and bases question

[nemonilla]

Homework Statement

Homework Equations

The Attempt at a Solution

i have answered the first part, explained why its a basis

i think that i need to show that the N(T) = V alongside R(T) = W will give matrices that cannot be multiplied therefore T is the one that can't exist
not sure though

and i think for the final part i just multiply such matrix but i don't know what to multiply?!

very confusing!

 

[lippyka]

Let V = R2, W = R3 and T:V → W be a linear transformation defined byT(x,y) = (x + y, 2x + y, x - y). Part 1: Prove that {(1, 0), (0, 1)} is a basis for V.To show that {(1, 0), (0, 1)} is a basis for V, we must show that it spans V and is linearly independent. To show that it spans V, we need to show that all elements in V can be written as linear combinations of (1, 0) and (0, 1). Let v = (a, b) be an arbitrary element of V. Then v = a(1, 0) + b(0, 1) = (a, 0) + (0, b) = (a + 0, 0 + b) = (a, b).This shows that any element of V can be written as a linear combination of (1, 0) and (0, 1), so the set {(1, 0), (0, 1)} spans V.To show that the set is linearly independent, we need to show that the only solution to the equation a(1, 0) + b(0, 1) = 0 is when a = b = 0. Substituting a = 0 and b = 0 into the equation gives:0(1, 0) + 0(0, 1) = 0 which is true, so the only solution is when a = b = 0. This shows that the set is linearly independent, so {(1, 0), (0, 1)} is a basis for V.Part 2: Show that there does not exist any linear transformation S: W → V such that S ◦ T = I.To show that there does not exist any linear transformation S: W → V such that S ◦ T = I, we need to show that the matrices N(T) and R(T) do not have a common multiple. Let N(T) = [1 1 2] and R(T) = [1 2 -1]. These matrices cannot be multiplied together because their number of columns (