235_course_notes_chapter1

# 235_course_notes_chapter1 - Chapter 1 Fundamental Subspaces...

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Chapter 1 Fundamental Subspaces The main purpose of this chapter is to review several important concepts from Linear Algebra 1. These concepts include subspaces, bases, dimension, and linear mappings. It is important to ensure that you are very familiar with these concepts as we will be using them, along with other Linear Algebra 1 concepts, a lot in Math 235. Note that the Math 136 course notes are provided on the course website for your reference. 1.1 Bases of Fundamental Subspaces DEFINITION Fundamental Subspaces Let A be an m × n matrix. The four fundamental subspaces of A are (i) The columnspace of A . Col( A ) = { Avx | vx R n } (ii) The rowspace of A . Row( A ) = { A T vx | vx R m } (iii) The nullspace of A . Null( A ) = { vx R n | Avx = v 0 } (iv) The left nullspace of A . Null( A T ) = { vx R m | A T vx = v 0 } It is easy to use the Subspace Test to show that Col( A ) and Null( A T ) are subspace of R m , and Row( A ) and Null( A ) are subspace of R m . Our goal now is to Fnd an easy way to determine a basis for each of the four fundamental subspaces. THEOREM 1 Let A be an m × n matrix. The columns of A which correspond to leading ones in the reduced row echelon form of A form a basis for Col( A ). Moreover, dim Col( A ) = rank( A ) Proof: We Frst observe that if A is the zero matrix, then the result is trivial. Hence, we can assume that rank( A ) = r > 0. 1

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2 Chapter 1 Fundamental Subspaces Denote the columns of the reduced row echelon form R of A by v r 1 , . . . ,v r n . Since rank( A ) = r , R contains r leading ones. Let t 1 , . . . , t r denote the indexes of the columns of R which contain leading ones. We will ±rst show that B = { v r t 1 , . . . ,v r t r } is a basis for Col( R ). Observe that by de±nition of the reduced row echelon form the vectors v r t 1 , . . . ,v r t r are standard basis vectors of R m and hence form a linearly independent set. Additionally, every column of R which does not contain a leading one can be written as a linear combination of the columns which do contain leading ones, so Span B = Col( R ). Therefore, B is a basis for Col( R ) as claimed. Denote the columns of A by va 1 , . . . ,va n . We will now show that C = { va t 1 , . . . ,va t r } is a basis for Col( A ) by using the fact that B is a basis for Col( R ). To do this, we ±rst need to ±nd a relationship between the vectors in B and C . Since R is the reduced row echelon form of A there exists a sequence of elementary matrices E 1 , . . . , E k such that E k ··· E 1 A = R . Let E = E k ··· E 1 . Recall that every elementary matrix is invertible, hence E - 1 = E - 1 1 ··· E - 1 k exists. Then R = EA = [ Eva 1 ··· Eva n ] Then v r i = Eva i , or va i = E - 1 v r i .
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## This note was uploaded on 11/28/2011 for the course MATH 235 taught by Professor Celmin during the Fall '08 term at Waterloo.

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235_course_notes_chapter1 - Chapter 1 Fundamental Subspaces...

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