Math 113 Midterm Solutions
نویسنده
چکیده
V = U ⊕ U ′, for some other subspace U ′ ⊂ V , by Theorem 3.21. Consider T restricted to U ′. This is a map T : U ′ →W , which is injective, since U ′∩ker(T ) = {0}. And note that injective maps, in general, preserve linear independence: if we have a linearly independent set {~x1, . . . , ~xn} ∈ X in a vector space X, a linear map of vector spaces A : X → Y , and a linear combination 0 = λ1A~x1 + · · ·+ λnA~xn, then 0 = A(λ1~x1 + · · ·+ λn~xn), so λ1~x1 + · · ·+ λn~xn = 0, which implies λ1 = · · · = λn = 0. Hence, choosing arbitrary bases for U and U ′ automatically gives us the desired properties.
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