By Edwards, Charles Henry

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**Example text**

V1–V8 are all immediate consequences of our definitions and the properties of . For example, to prove V6, let x = (x1, . . , xn). Then The remaining verifications are left as exercises for the student. A vector space is a set V together with two mappings V × V → V and × V → V, called vector addition and scalar multiplication respectively, such that V1–V8 above hold for all and (V3 asserts that there exists such that x + 0 = x for all , and V4 that, given , there exists such that x + (− x) = 0).

Vn for V. Since it is clear (by linearity of L) that the vectors L( v1), . . , L( vn) generate Im L, it suffices to prove that they are linearly independent. Suppose Then so t1 v1 + · · · + tn vn = 0 because Ker L = 0. But then t1 = · · · = tn = 0 because the vectors v1, . . , vn are linearly independent. 3). 2 Let L : n → m be defined by L( x) = A x, where A = (aij) is an m × n matrix. Then (a) Ker L is the orthogonal complement of that subspace of n that is generated by the row vectors A1, .

3). 2 Let L : n → m be defined by L( x) = A x, where A = (aij) is an m × n matrix. Then (a) Ker L is the orthogonal complement of that subspace of n that is generated by the row vectors A1, . . , Am of A, and (b) Im L is the subspace of m that is generated by the column vectors A1, . . , An of A. PROOF (a) follows immediately from the fact that L is described by the scalar equations so that the ith coordinate Li( x) is zero if and only if x is orthogonal to the row vector Ai. (b) follows immediately from the fact that Im L is generated by the images L( e1), .