By William C. Brown

This textbook for senior undergraduate and primary yr graduate-level classes in linear algebra and research, covers linear algebra, multilinear algebra, canonical varieties of matrices, general linear vector areas and internal product areas. those issues supply the entire must haves for graduate scholars in arithmetic to arrange for advanced-level paintings in such parts as algebra, research, topology and utilized mathematics.

Presents a proper method of complex issues in linear algebra, the math being awarded essentially by way of theorems and proofs. Covers multilinear algebra, together with tensor items and their functorial houses. Discusses minimum and attribute polynomials, eigenvalues and eigenvectors, canonical kinds of matrices, together with the Jordan, actual Jordan, and rational canonical kinds. Covers normed linear vector areas, together with Banach areas. Discusses product areas, overlaying genuine internal product areas, self-adjoint differences, complicated internal product areas, and common operators.

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LINEAR ALGEBRA DUALS AND ADJOINTS Let V be a vector space over F. 1: V* = HomF(V, F) is called the dual of V. 25 that V* is finite dimensional with dim V* = dim V. 2. 2: Let V be finite dimensional. Then dim V* = dim V. Proof Let ocj be a basis of V. , n, define an = P is the isomorphism determined element xr e V* by czr = by the basis and P F is the natural projection onto the ith coordinate of P. Thus, if x = + + Here V —. then xr is given by cxt(x1cz1 + ... + = x1 = 0. Let is a basis of V*. Suppose We claim that z* = .

20: F" a)' is a commutative diagram of isomorphisms. 20 is commutative. 15. We need only argue M(b, a) is an -÷ F" is an isomorphism. invertible matrix. Then the map M(b, a)t = SM(La): Now change of basis matrices M(ö, a) are always invertible. 18. For any fJeV, we have M(a, b)M(ö, a)[fl]5 = M(a, b)[fl]a = [fl]5. This equation easily implies M(a, ö)M(ö, a) = I,,, the n x n identity matrix. 6. 21: In this example, we construct a vector space isomorphic to V". Let V be a vector space over F, and let n e N.

16 commutes. Only the uniqueness of I remains to be proved. If T' e Hom(V/W, V') is another map for which T'H = T, then I = T' on Im H. But H is surjective. Therefore, I = T'. 17: Suppose T e V'). Then Tm T V/ker T. Proof We can view T as a surjective, linear transformation from V to Tm T. 18, H is the natural map from V to V/ker T. We claim I is an isomorphism. Since IH = T and T: V —+ Tm T is surjective, I is surjective. Suppose & e ker I. Then T(cz) = TH(cz) = 1(ä) = 0. Thus, e ker T. But, then fl(cz) = 0.