This problem provides a useful way to construct invertible linear maps between subspaces X ⊂ Rn and Y ⊂ Rm.3 Let B = {u1, u2, · · · , uk} be a basis of X. (a) Let {v1, v2, · · · , vk} ⊂ Y . Show that there exists a unique linear transform T : X → Y such that T(uj) = vj for every 1 ≤ j ≤ k. To be specific, you need to 1) define a function4 T : X → Y and verify that it is linear, 2) show that if T0 : X → Y is a linear map such that T0(uj) = vj, then T = T0.

To solve this problem, we need to follow these steps:

1. Define a function T: X → Y:

We can define a function T: X → Y such that for any x in X, we can write x as a linear combination of the basis vectors in B. That is, x = c1u1 + c2u2 + ... + ckuk for some scalars c1, c2, ..., ck. We then define T(x) = c1v1 + c2v2 + ... + ckvk.

2. Verify that T is linear:

To show that T is linear, we need to show that for any x, y in X and any scalar a, T(x + y) = T(x) + T(y) and T(ax) = aT(x).

Given x = c1u1 + c2u2 + ... + ckuk and y = d1u1 + d2u2 + ... + dkuk, we have T(x + y) = T((c1 + d1)u1 + (c2 + d2)u2 + ... + (ck + dk)uk) = (c1 + d1)v1 + (c2 + d2)v2 + ... + (ck + dk)vk = c1v1 + c2v2 + ... + ckvk + d1v1 + d2v2 + ... + dkvk = T(x) + T(y).

Given x = c1u1 + c2u2 + ... + ckuk, we have T(ax) = T(ac1u1 + ac2u2 + ... + ackuk) = ac1v1 + ac2v2 + ... + ackvk = a(c1v1 + c2v2 + ... + ck*vk) = aT(x).

So T is linear.

3. Show that if T0 : X → Y is a linear map such that T0(uj) = vj, then T = T0:

Suppose T0 is another linear map such that T0(uj) = vj for every 1 ≤ j ≤ k. Then for any x in X, we can write x = c1u1 + c2u2 + ... + ckuk, and we have T0(x) = T0(c1u1 + c2u2 + ... + ckuk) = c1T0(u1) + c2T0(u2) + ... + ckT0(uk) = c1v1 + c2v2 + ... + ckvk = T(x).

So T = T0.

Therefore, there exists a unique linear transform T : X → Y such that T(uj) = vj for every 1 ≤ j ≤ k.