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Show that the function T: V → W as defined in our proof of Theorem 10 (§6.2 class notes) is a linear transformation.

Show that the function T: V → W as defined in our proof of Theorem 10 (§6.2 class notes) is a linear transformation.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question 3, and the first three slides supporting theorem 10 for the question

Q3 Show that the function T: V → W as defined in our proof of Theorem 10 (§6.2 class notes) is a linear transformation.
Transcribed Image Text:Q3 Show that the function T: V → W as defined in our proof of Theorem 10 (§6.2 class notes) is a linear transformation.
Two finite-dimensional vector spaces are isomorphic if and only if they have the same dimension. proof: (Let V be a vector space with dim (v) = n, and suppose vew. Then there is a linear transformation TV-W that is one-to-one and onto. Since T is one-to-one, then nullity (T) = 0 So by the rank-nulity theorem, rank (T) = dim (v) - nullity (T) = n - 0 = n. Since T is onto, dim(w) = rank (T) () Suppose dim(v) = dim (w) - n. Take a basis B = { V₁, V₂, ... Vn² of V B' = { W₁, W₂,..., Winb of W. = n. Thus, dim(v) =dim(w) TLV) = TGV₁+ C₂v₂ + + = and If VEV, then we can find unique scalars c₁, c₂...., CnER SO that = C₁V₁ + C₂ √₂ + Cnn. Define T: VW by (nVn) ++ C₁ W₁+CaNa +---+ cnwn. Defining T this way yields a linear transformation. a To see that T is one-to-one, we check kerCT). Suppose V=C₁V₁ +₂√√₂+ + (nun & ker (T). Then, = T(V) = C₁W₁ + C₂ W₂ + .-- + Cn Wn. Since B' is a basis, it is linearly independent, so G₁ = Ca = Cn=0. Thus, V=Ov₁ + O₂ + ㅎㅎㅎ.. ...+ 0√₁ = 0 + Thus, ker (T) = {of and I is one-to-one. ++ basis
Transcribed Image Text:Two finite-dimensional vector spaces are isomorphic if and only if they have the same dimension. proof: (Let V be a vector space with dim (v) = n, and suppose vew. Then there is a linear transformation TV-W that is one-to-one and onto. Since T is one-to-one, then nullity (T) = 0 So by the rank-nulity theorem, rank (T) = dim (v) - nullity (T) = n - 0 = n. Since T is onto, dim(w) = rank (T) () Suppose dim(v) = dim (w) - n. Take a basis B = { V₁, V₂, ... Vn² of V B' = { W₁, W₂,..., Winb of W. = n. Thus, dim(v) =dim(w) TLV) = TGV₁+ C₂v₂ + + = and If VEV, then we can find unique scalars c₁, c₂...., CnER SO that = C₁V₁ + C₂ √₂ + Cnn. Define T: VW by (nVn) ++ C₁ W₁+CaNa +---+ cnwn. Defining T this way yields a linear transformation. a To see that T is one-to-one, we check kerCT). Suppose V=C₁V₁ +₂√√₂+ + (nun & ker (T). Then, = T(V) = C₁W₁ + C₂ W₂ + .-- + Cn Wn. Since B' is a basis, it is linearly independent, so G₁ = Ca = Cn=0. Thus, V=Ov₁ + O₂ + ㅎㅎㅎ.. ...+ 0√₁ = 0 + Thus, ker (T) = {of and I is one-to-one. ++ basis
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