Proof homework due Class 26

Proof the theorem about linear transformations of arbitrary vector spaces.

Theorem.

1. If f is a linear transformation from V to W then f(0)=0.

2. If f is a linear transformation from V to W and g is a linear transformation from W to Y (V, W, Y are vector spaces) then the product (composition) gf is a linear transformation from V to Y.

3. If f and g are linear transformations from V to W (V and W are vector spaces) then the sum f+g which takes every vector A in V to the sum f(A)+g(A) in W is again a linear transformation from V to W.

4. If f is a linear transformation from V to W and k is a scalar then the map kf which takes every vector A in V to k times f(A) is again a linear transformation from V to W.

Part 1

By the characterization of linear transformations, we know that:

A function f from Rn to Rm is a linear transformation if and only if it satisfies the following two properties:

  1. For every two vectors A and B in Rn

    f(A+B)=f(A)+f(B);

  2. For every vector A in Rn and every number k

    f(kA)=kf(A).

Thus for every B in V, f(0+B) = f(0) + f(B). Since 0+B = B, this says f(B) = f(0) + f(B). Subtracting f(B) from both sides yields f(0) = 0.

Part 2.

If f is a linear tranformation from V to W and g is a linear transformation from W to Y (V,W,Y are vector spaces) then the product (composition) gf is a linear transformation from V to Y.

If f and g are linear, A and B are vectors from V, and k is any scalar, then by the properties of linear transformations,

gf(A+B) = g(f(A+B)) = g(f(A) + f(B)) = g(f(A)) + g(f(B)) = gf(A) + gf(B).

So, the transformation gf obeys the first property of linear transformations. We can also say,

gf(kA) = g(f(kA)) = g(k*f(A)) = k*g(f(A)) = k*gf(A).

Hence, the transformation gf also obeys the second property of linear transformations. The proof of this property is complete.

Part 3.

Once again, we are asked to show that something is indeed a linear transformation. Refering to the characterization of linear transformations, we see that this is true if and only if for every two vectors A and B in Rn

f(A+B)=f(A)+f(B);

And for every vector A in Rn and every number k

f(kA)=kf(A).

So, let us denote the sum f+g as h. For arbitrary vectors A and B, h(A+B) = f(A+B) + g(A+B) = f(A) + f(B) + g(A) + g(B) = (f(A) + g(A)) + (f(B) + g(B)) = (f+g)(A) + (f+g)(B) = h(A) + h(B)

Again for the arbitrary vector A, h(kA) = f(kA) + g(kA) = kf(A) + kg(A) = k(f(A) + g(A)) = k((f+g)(A)) = kh(A)

Part 4.

Similarly to Part 3, we once again invoke the power of the characterization of linear transformations. Here, let h be kf. Then for our arbitrary vectors A and B, h(A+B) = kf(A + B) = kf(A) + kf(B) = h(A) + h(B)

For the second half of this, let us use an arbitrary constant c, to avoid confusion with our already-in-use k.
So, h(cA) = kf(cA) = c(kf(A)) =c(h(A)) = ch(A)

Thus, with all four parts proved, the proof is complete.