By Spencer, Donald Clayton; Nickerson, Helen Kelsall

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Isomorphism of vector spaces Proposition. A linear transformation is injective i f and only i f ker T = Proof. Since 'av = ker T. Conversely, let V ~-> W 'av· T is linear, jective, only one vector of v T can have 'av e 'aw ker T. If T is in- as an image, so T be linear and such that 39 ker T = 'av· T(A) = T(A' ). A = A'. 2. v are vectors of V such that Suppose A, A' A - A' e ker T = 1 t\,; hence T is injective. Proposition. Let T V -> W be linear, where are finite dimensional. (i) If (ii) If T is surjective, then (iii) If w.

By hypothesis there is a finite set D' vectors such that and dim ~ k. 5 asserts that k ~ h. and the proof is complete. A1, •.. , Ak is a basis for V, and then there is one and only one sequence of coefficients •I xk Theorem. If k such that Proof. A= Ei= 1xiAi. Since A£ L{A 1, ••• , Ak)' there is at least k k one such representatipn. Suppose A = Ei=txiAl and A = Ei=lyiAi are two representations. Then E~= 1 (xi - yi)Ai = ~. Since A1, ••• , Ak are independent, each coefficient must be zero. Therefore xi = Yi for each i, and the representation is unique.

Tions by constants, they have essentially the same properties. These will be developed in the exercises. §3. 1. Exercises Let V and W be sets, and let with domain V and range w. Let D and T be a. function E be subsets of v. ) T (D U E) = T(D) U T (E) , (b) T (D n E) Let D' and E' ( c) 2. from V to such that (i) that Let w. n T (E) • be subsets of w. Show that T- 1 (D' u EI ) T- 1 (D 1 ) U T- 1 (E') T~ 1 (D'n E') (d) c T(D) T- 1 (D 1 ) n T- 1 (E 1 ) V and W be sets, and let Suppose there exists a.