4.3 Help - Unit 22 Basis and Dimension In Unit 20, we...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Unit 22 Basis and Dimension In Unit 20, we studied the concept of a set of vectors being linearly inde- pendent or linearly dependent. In Unit 21, we learned about subspaces and spanning sets of subspaces. We now bring these ideas together to get the concept of a basis for a subspace, which allows us to define/determine the dimension of the subspace. Definition 22.1. Let S be any subspace of Rfractur m . A set T of vectors in S is said to be a basis for S if both of the following conditions are met: 1. T spans S . 2. T is a linearly independent set. Example 1 . Show that T = { vectorv 1 ,vectorv 2 ,vectorv 3 } , with vectorv 1 = (1 , 1 , 1), vectorv 2 = (1 , 2 , 3) and vectorv 3 = (- 1 , , 2) is a basis for Rfractur 3 . Solution: We must show that the set T is linearly independent and spans Rfractur 3 . Let vectorx = ( x 1 , x 2 , x 3 ) be an arbitrary vector in Rfractur 3 . For T to span Rfractur 3 , it must be possible to write vectorx as a linear combination of the vectors in T . That is, we must be able to find constants c 1 , c 2 , and c 3 such that c 1 (1 , 1 , 1) + c 2 (1 , 2 , 3) + c 3 (- 1 , , 2) = ( x 1 , x 2 , x 3 ) Carrying through the vector arithmetic and equating corresponding com- ponents, we get the SLE c 1 + c 2- c 3 = x 1 c 1 + 2 c 2 = x 2 c 1 + 3 c 2 + 2 c 3 = x 3 1 We must show that this system has at least one solution. The coefficient matrix is: A = 1 1- 1 1 2 1 3 2 The RREF of A is I 3 , which we know means that the system has a unique solution. That is, there are such scalars and vectorx can be written as a linear combination of the vectors in T . Therefore T spans Rfractur 3 . Furthermore, since the columns of A are the vectors in the set T and the RREF of A is I 3 (which of course contains only distinct elementary columns), then by Theorem 20.6, the set T is linearly independent. Thus we see that T is a linearly independent set of vectors from Rfractur 3 which also spans Rfractur 3 , so T is a basis for Rfractur 3 . Example 2 . Show that the set containing the elementary vectors of Rfractur m is a basis for Rfractur m . Solution: We have seen (in Unit 21, Example 5) that the set of elementary vectors in Rfractur m spans Rfractur m . But clearly this set is linearly independent. That is, if we form the matrix whose columns are the vectors vectore 1 , vectore 2 , ... , vectore m , what we have is the m m identity matrix, I m , so the matrix is in RREF and all of the columns are (distinct) elementary columns. Thus we see that the set of elementary vectors in Rfractur m is a subset of Rfractur m which both spans Rfractur m and is linearly independent, and therefore is a basis for Rfractur m . Suppose now that we start with some spanning set T of a subspace S and wish to trim the set until it is linearly independent, in order to obtain a basis for S . Two questions arise: Is it always possible to do this? If so, how? The answer to the first question is yes, as the following Theorem tells us. The second question is answered by the subsequent procedure.second question is answered by the subsequent procedure....
View Full Document

This note was uploaded on 04/17/2008 for the course MAT MAT187 taught by Professor Chulkov during the Fall '07 term at University of Toronto- Toronto.

Page1 / 10

4.3 Help - Unit 22 Basis and Dimension In Unit 20, we...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online