I am reading (theorem 2.14) from a textbook, and don't understand how g = Tf and (#1) line of reasoning. The theorem and proof is as follows:(adsbygoogle = window.adsbygoogle || []).push({});

Theorem 2.14: Let V and W be finite-dimensional vector spaces having ordered bases B and C, respectively, and let T: V-->W be linear. Then, for each u in V, we have

[T(u)]_{C}= [T]_{B}^{C}_{B}.

Proof: Fix u in V, and define the linear transformations f: F --> V by f(a) = au and

g: F-->W by g(a) = aT(u) for all a in F. Let A = {1} be the standard

ordered basis for F. Notice that g = Tf.. Identifying

column vectors as matrices and using Theorem 2.11, we obtain

(#1) [T(u)]_{C}= [g(1)]_{C}= [g]_{A}^{C}= [Tf]_{A}^{C}= [T]_{B}^{C}[f]_{A}^{B}= [T]_{B}^{C}[f(1)]_{B}= [T]_{B}^{C}_{B}.

-------------------------------

Theorem 2.11: Let V, W, and Z be finite-dimensional vector spaces with ordered bases, A,B, and C, respectively. Let T: V-->W and U: W-->Z be linear transformations. Then

[UT]_{A}^{C}=_{B}^{C}[T]_{A}^{B}

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# Composition of Linear Transformation and Matrix Multiplication

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