Download Aus dem Nachlass von R. Brauer by Jørn Børling Olsson PDF

By Jørn Børling Olsson

Show description

Read Online or Download Aus dem Nachlass von R. Brauer PDF

Best algebra books

Algebra: A Text-Book of Determinants, Matrices, and Algebraic Forms

A number of the earliest books, fairly these relationship again to the 1900s and earlier than, at the moment are tremendous scarce and more and more pricey. we're republishing those vintage works in reasonable, top of the range, glossy variants, utilizing the unique textual content and paintings.

Extra info for Aus dem Nachlass von R. Brauer

Sample text

F/. Let m denote the highest degree of the polynomials in this list. Then every polynomial in the span of this list has degree at most m. Thus z mC1 is not in the span of our list. F/. F/ is infinite-dimensional. v1 ; : : : ; vm /. By the definition of span, there exist a1 ; : : : ; am 2 F such that v D a1 v1 C C am vm : Consider the question of whether the choice of scalars in the equation above is unique. v1 ; : : : ; vm /. If the only way to do this is the obvious way (using 0 for all scalars), then each aj cj equals 0, which means that each aj equals cj (and thus the choice of scalars was indeed unique).

In general, a vector space is an abstract entity whose elements might be lists, functions, or weird objects. B Definition of Vector Space 15 Soon we will see further examples of vector spaces, but first we need to develop some of the elementary properties of vector spaces. The definition of a vector space requires that it have an additive identity. The result below states that this identity is unique. 25 Unique additive identity A vector space has a unique additive identity. Proof Suppose 0 and 00 are both additive identities for some vector space V.

To show that U \ W D f0g, suppose v 2 U \ W. Then there exist scalars a1 ; : : : ; am ; b1 ; : : : ; bn 2 F such that v D a1 u1 C C am um D b1 w1 C C bn wn : Thus a1 u1 C C am u m b1 w1 bn wn D 0: Because u1 ; : : : ; um ; w1 ; : : : ; wn is linearly independent, this implies that a1 D D am D b 1 D D bn D 0. Thus v D 0, completing the proof that U \ W D f0g. B 1 Find all vector spaces that have exactly one basis. 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 .

Download PDF sample

Rated 4.28 of 5 – based on 16 votes