By T. S. Blyth

This textbook presents a self-contained path at the uncomplicated homes of modules and their significance within the thought of linear algebra. the 1st eleven chapters introduce the important effects and purposes of the idea of modules. next chapters care for complicated linear algebra, together with multilinear and tensor algebra, and discover such themes because the external product method of the determinants of matrices, a module-theoretic method of the constitution of finitely generated Abelian teams, canonical types, and common adjustments. compatible for undergraduate classes, the textual content now contains a facts of the distinguished Wedderburn-Artin theorem which determines the constitution of straightforward Artinian earrings.

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In particular, every n-dimensional vector space V is isomorphic to F n . Proof. T // Ä m < n so T is not an epimorphism. T // m n > 0 so T is not a monomorphism. (3) Parts (1) and (2) show that if m ¤ n, then V and W are not isomorphic. If m D n, choose a basis fv1 ; : : : ; vm g of V and a basis fw1; : : : ; wm g of W . 24 T is an isomorphism. ✐ ✐ ✐ ✐ ✐ ✐ “book” — 2011/3/4 — 17:06 — page 20 — #34 ✐ ✐ 20 1. 7. Let A be an n-by-n matrix. The following are equivalent: (1) A is invertible. (10 ) The equation Ax D b has a unique solution for every b 2 F n .

Let A be an n-by-n matrix. The following are equivalent: (1) A is invertible. (10 ) The equation Ax D b has a unique solution for every b 2 F n . (2) The equation Ax D b has a solution for every b 2 F n . (3) The equation Ax D 0 has only the trivial solution x D 0. Proof. 2 into matrix language. We emphasize that this one-sentence proof is the “right” proof of the equivalence of these properties. For the reader who would like to see a more computational proof, we shall prove directly that (1) and (10 ) are equivalent.

Y / D 0, then y D 0. v/ D 0 for every v 2 V . v// for every v 2 V . Choose x 2 X. v// D 0. , y D 0. Suppose that T is not surjective. T / is a proper subspace of X. T / and extend this set to a basis C D fx1 ; x2; : : : ; x10 ; x20 ; : : :g of X. xj0 / D 0 for j ¤ 1. T /. y / D 0. (3) This immediately follows from (1) and (2). Next we see how the dual behaves under composition. 6. 23. Let T W V ! W and S W W ! X be linear transformations. Then S ı T W V ! S ı T / W X ! S ı T / D T ı S . Proof. Let y 2 X and let x 2 X.