To show that the set of linear transformations T ∈ ζ(V,W) such that T is not surjective is not a subspace of ζ(V,W), we need to show that it fails at least one of the subspace properties: closed under addition or closed under scalar multiplication.
Let's assume by contradiction that the set S = {T ∈ ζ(V,W) : T is not surjective} is a subspace of ζ(V,W).
1. Closure under addition:
Let T1 and T2 be two linear transformations in S that are not surjective. Since S is assumed to be a subspace, their sum T = T1 + T2 should also be in S. We want to show that T is not surjective.
If T is surjective, then for every w ∈ W, there exists v ∈ V such that T(v) = w. Since T1 and T2 are not surjective, there exist some w1, w2 ∈ W such that there is no v1 ∈ V with T1(v1) = w1 and no v2 ∈ V with T2(v2) = w2.
Now, consider w = w1 + w2. If T is surjective, there should exist v ∈ V such that T(v) = w. However, this implies that T1(v) + T2(v) = w, which means T1(v) = w1 and T2(v) = w2, contradicting our assumption. Therefore, T cannot be surjective, which means S fails to be closed under addition.
2. Closure under scalar multiplication:
Let T be a linear transformation in S and consider α to be a scalar. We want to show that αT is not surjective.
Suppose, for the sake of contradiction, that αT is surjective. Then, for every w ∈ W, there exists v ∈ V such that αT(v) = w. However, this implies that T(v) = α^(-1)w.
Now, if T is not surjective, there exists some w ∈ W such that there is no v ∈ V with T(v) = w. Let's choose such a w and consider α^(-1)w. Since T is not surjective, there is no v ∈ V with T(v) = α^(-1)w. But this contradicts our assumption that αT is surjective.
Therefore, αT cannot be surjective, and S fails to be closed under scalar multiplication.
Since S fails to satisfy both closure under addition and closure under scalar multiplication, we can conclude that the set of linear transformations T ∈ ζ(V,W) such that T is not surjective is not a subspace of ζ(V,W).
Let's assume by contradiction that the set S = {T ∈ ζ(V,W) : T is not surjective} is a subspace of ζ(V,W).
1. Closure under addition:
Let T1 and T2 be two linear transformations in S that are not surjective. Since S is assumed to be a subspace, their sum T = T1 + T2 should also be in S. We want to show that T is not surjective.
If T is surjective, then for every w ∈ W, there exists v ∈ V such that T(v) = w. Since T1 and T2 are not surjective, there exist some w1, w2 ∈ W such that there is no v1 ∈ V with T1(v1) = w1 and no v2 ∈ V with T2(v2) = w2.
Now, consider w = w1 + w2. If T is surjective, there should exist v ∈ V such that T(v) = w. However, this implies that T1(v) + T2(v) = w, which means T1(v) = w1 and T2(v) = w2, contradicting our assumption. Therefore, T cannot be surjective, which means S fails to be closed under addition.
2. Closure under scalar multiplication:
Let T be a linear transformation in S and consider α to be a scalar. We want to show that αT is not surjective.
Suppose, for the sake of contradiction, that αT is surjective. Then, for every w ∈ W, there exists v ∈ V such that αT(v) = w. However, this implies that T(v) = α^(-1)w.
Now, if T is not surjective, there exists some w ∈ W such that there is no v ∈ V with T(v) = w. Let's choose such a w and consider α^(-1)w. Since T is not surjective, there is no v ∈ V with T(v) = α^(-1)w. But this contradicts our assumption that αT is surjective.
Therefore, αT cannot be surjective, and S fails to be closed under scalar multiplication.
Since S fails to satisfy both closure under addition and closure under scalar multiplication, we can conclude that the set of linear transformations T ∈ ζ(V,W) such that T is not surjective is not a subspace of ζ(V,W).
