By I. N. Herstein

Starting summary Algebra with the vintage Herstein therapy.

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Mathematik für Wirtschaftswissenschaftler 2: Lineare Algebra, Funktionen mehrerer Variablen

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28. x1 ; x2 ; x3 ; x4 ; x5 / 2 R5 W x1 D 3x2 and x3 D 7x4 g: Find a basis of U. 4 (b) Extend the basis in part (a) to a basis of R5 . (c) Find a subspace W of R5 such that R5 D U ˚ W. z1 ; z2 ; z3 ; z4 ; z5 / 2 C5 W 6z1 D z2 and z3 C2z4 C3z5 D 0g: Find a basis of U. (b) Extend the basis in part (a) to a basis of C5 . (c) Find a subspace W of C5 such that C5 D U ˚ W. F/ such that none of the polynomials p0 ; p1 ; p2 ; p3 has degree 2. 6 Suppose v1 ; v2 ; v3 ; v4 is a basis of V. Prove that v1 C v2 ; v2 C v3 ; v3 C v4 ; v4 is also a basis of V.

32). Of course u1 ; : : : ; um is a linearly independent list of vectors in V. 33). w1 ; : : : ; wn /. 45 we need only show that Proof V DU CW and U \ W D f0g: To prove the first equation above, suppose v 2 V. Then, because the list u1 ; : : : ; um ; w1 ; : : : ; wn spans V, there exist a1 ; : : : ; am ; b1 ; : : : ; bn 2 F such that v D a1 u1 C C am um C b1 w1 C C bn wn : „ ƒ‚ … „ ƒ‚ … u w In other words, we have v D u C w, where u 2 U and w 2 W are defined as above. Thus v 2 U C W, completing the proof that V D U C W.

Show that RR D Ue ˚ Uo . CHAPTER 2 American mathematician Paul Halmos (1916–2006), who in 1942 published the first modern linear algebra book. The title of Halmos’s book was the same as the title of this chapter. 1 Notation F, V F denotes R or C. V denotes a vector space over F. In the last chapter we learned about vector spaces. Linear algebra focuses not on arbitrary vector spaces, but on finite-dimensional vector spaces, which we introduce in this chapter. LEARNING OBJECTIVES FOR THIS CHAPTER span linear independence bases dimension © Springer International Publishing 2015 S.

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