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By Spencer, Donald Clayton; Nickerson, Helen Kelsall

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For Axiom G2, note that the identity therefore an element of A(V). therefore an element of A(V). 4 to conclude that the inverse function T- 1 is an isomorphism and Remark. Let We. can "divide" in E(V) T be an automorphism, and suppose in the following sense. s 1, S2 e E(V) are such that TS 1 = S2 • Multiply both sides by T- 1 on the left. 2, we obtain T- 1 (Ts, ) = (T- 1T )s, = that is, we have solved for solved for s 1 = S2T- 1 • s 1 = T-1 s 2 . rs, = s1 ; Similarly S1T = s 2 can Thus the usual rule "one can divide by any- thing different from zero", must be replaced by "one can divide, on the left or on the right, by any automorphism".

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. dim v ':?. dim w. dim v = dim w. T is injective, then dim V T is bijective, then Proof. 3. and T : V = Theorem.

1. vector space V, Quotient; direct sum Definition. 6) and for which the operations of modulo addition and scalar multiplication are defined by the condition that the function j V -> V/U, which assigns to each A e V the equivalence class in which it lies, be a linear transformation. Proof. that B = j The properties of the equivalence relation show is surjective and that A + X for some vector space such that (1) j(A) + j(B) u. 1. for all It A v, x e e R , remains to verify that these conditions.

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