Math 108a Hw 6 Solutions

‘⇒’ Let V be a finite dimensional vector space and let T ∈ L(V, W ). Assume that T is surjective. Let B = {v1, . . . vn} be a basis for V . Notice that even though a priori W need not be finite dimensional, the fact that T is surjective implies that T (B) = {T (v1), . . . T (vn)} is a spanning set for W and thus W is finite dimensional. Since T (B) spans W (a finite dimensional vector space), we can reduce it to a basis, B′, for W . We will denote B′ = {w1, . . . wk} where for each wi ∈ B′, wi = T (vj) for some vj ∈ B. Since for each wi ∈ B′, there exists at least one (there might be more, but that won’t matter) vj ∈ B such that T (vj) = wi, we will define S(wi) to be vj (i.e. vj is one of the elements of B such that T (vj) = wi). Since S has been defined on B′, a basis of W , S ∈ L(W, V ). Thus T ◦ S ∈ L(W, W ) = L(W ). To show that T ◦ S = IW , the identity linear map on W , it is enough to show that for each wi ∈ B′, S ◦ T (wi) = wi. So let wi ∈ B′. T ◦ S(wi) = T (vj), where vj is an element in B such that T (vj) = wi. Thus T ◦ S(wi) = T (vj) = wi. Thus T ◦ S = IW and we are done. ‘⇐’ Let V be a finite dimensional vector space and let T ∈ L(V, W ). Assume that there exists S ∈ L(W, V ) such that T ◦ S = IW , the identity map on W . Assume for contradiction that T is not surjective. Thus, there exists w ∈ W such that w / ∈ T (V ), the image of V . T ◦ S(w) = IW (w) = w. Thus w ∈ T (S(W )). However, since S(W ) ⊂ V (it is actually a linear subspace of V ), we know that T (S(W )) ⊂ T (V ). But then w ∈ T (S(W )) ⇒ w ∈ T (V ) which contradicts the assumption that w / ∈ T (V ) and thus it is a contradiction to assume that such a w exists. Thus ∀w ∈ W , there exists some v ∈ V such that T (v) = w and ergo, T is