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jeff1evesque
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Theorem 2.15:
Let A be an m x n matrix with entries from F. Then the left-multiplication transformation
[tex]L_A: F^n --> F^m[/tex]. Furthermore, if B is any other m x n matrix ( with entries from F ) and B and D are the standard ordered bases for [tex]F^n and F^m[/tex], respectively, then we have the following properties.
(d.) If [tex]T: F^n --> F^m[/tex] is linear, then there exists a unique m x n matrix C such that [tex]T = L_C[/tex]. In fact [tex]C = [T]_B ^D[/tex]
proof: Let [tex] C = [T]_B ^D[/tex]. By Theorem 2.14, we have [tex][T(x)]_D = [T]_B ^D[x]_B[/tex] or T(x) = Cx = [tex]L_C(x)[/tex] for all x in [tex]F^n[/tex]. So T = [tex]L_C[/tex]
In particular I don't understand how T(x) = CxThanks,JL
Let A be an m x n matrix with entries from F. Then the left-multiplication transformation
[tex]L_A: F^n --> F^m[/tex]. Furthermore, if B is any other m x n matrix ( with entries from F ) and B and D are the standard ordered bases for [tex]F^n and F^m[/tex], respectively, then we have the following properties.
(d.) If [tex]T: F^n --> F^m[/tex] is linear, then there exists a unique m x n matrix C such that [tex]T = L_C[/tex]. In fact [tex]C = [T]_B ^D[/tex]
proof: Let [tex] C = [T]_B ^D[/tex]. By Theorem 2.14, we have [tex][T(x)]_D = [T]_B ^D[x]_B[/tex] or T(x) = Cx = [tex]L_C(x)[/tex] for all x in [tex]F^n[/tex]. So T = [tex]L_C[/tex]
In particular I don't understand how T(x) = CxThanks,JL