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ma265 lecture notes

# ma265 lecture notes - Chapter 3 SUBSPACES 3.1 Introduction...

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Chapter 3 SUBSPACES 3.1 Introduction Throughout this chapter, we will be studying F n , the set of all n –dimensional column vectors with components from a field F . We continue our study of matrices by considering an important class of subsets of F n called subspaces . These arise naturally for example, when we solve a system of m linear ho- mogeneous equations in n unknowns. We also study the concept of linear dependence of a family of vectors. This was introduced briefly in Chapter 2, Remark 2.5.4. Other topics dis- cussed are the row space, column space and null space of a matrix over F , the dimension of a subspace, particular examples of the latter being the rank and nullity of a matrix. 3.2 Subspaces of F n DEFINITION 3.2.1 A subset S of F n is called a subspace of F n if 1. The zero vector belongs to S ; (that is, 0 S ); 2. If u S and v S , then u + v S ; ( S is said to be closed under vector addition); 3. If u S and t F , then tu S ; ( S is said to be closed under scalar multiplication). EXAMPLE 3.2.1 Let A M m × n ( F ). Then the set of vectors X F n satisfying AX = 0 is a subspace of F n called the null space of A and is denoted here by N ( A ). (It is sometimes called the solution space of A .) 55

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56 CHAPTER 3. SUBSPACES Proof. (1) A 0 = 0, so 0 N ( A ); (2) If X, Y N ( A ), then AX = 0 and AY = 0, so A ( X + Y ) = AX + AY = 0 + 0 = 0 and so X + Y N ( A ); (3) If X N ( A ) and t F , then A ( tX ) = t ( AX ) = t 0 = 0, so tX N ( A ). For example, if A = 1 0 0 1 , then N ( A ) = { 0 } , the set consisting of just the zero vector. If A = 1 2 2 4 , then N ( A ) is the set of all scalar multiples of [ - 2 , 1] t . EXAMPLE 3.2.2 Let X 1 , . . . , X m F n . Then the set consisting of all linear combinations x 1 X 1 + · · · + x m X m , where x 1 , . . . , x m F , is a sub- space of F n . This subspace is called the subspace spanned or generated by X 1 , . . . , X m and is denoted here by h X 1 , . . . , X m i . We also call X 1 , . . . , X m a spanning family for S = h X 1 , . . . , X m i . Proof. (1) 0 = 0 X 1 + · · · + 0 X m , so 0 ∈ h X 1 , . . . , X m i ; (2) If X, Y h X 1 , . . . , X m i , then X = x 1 X 1 + · · · + x m X m and Y = y 1 X 1 + · · · + y m X m , so X + Y = ( x 1 X 1 + · · · + x m X m ) + ( y 1 X 1 + · · · + y m X m ) = ( x 1 + y 1 ) X 1 + · · · + ( x m + y m ) X m ∈ h X 1 , . . . , X m i . (3) If X ∈ h X 1 , . . . , X m i and t F , then X = x 1 X 1 + · · · + x m X m tX = t ( x 1 X 1 + · · · + x m X m ) = ( tx 1 ) X 1 + · · · + ( tx m ) X m ∈ h X 1 , . . . , X m i . For example, if A M m × n ( F ), the subspace generated by the columns of A is an important subspace of F m and is called the column space of A . The column space of A is denoted here by C ( A ). Also the subspace generated by the rows of A is a subspace of F n and is called the row space of A and is denoted by R ( A ). EXAMPLE 3.2.3 For example F n = h E 1 , . . . , E n i , where E 1 , . . . , E n are the n –dimensional unit vectors. For if X = [ x 1 , . . . , x n ] t F n , then X = x 1 E 1 + · · · + x n E n . EXAMPLE 3.2.4 Find a spanning family for the subspace S of R 3 defined by the equation 2 x - 3 y + 5 z = 0.
3.2. SUBSPACES OF F N 57 Solution. ( S is in fact the null space of [2 , - 3 , 5], so S is indeed a subspace of R 3 .) If [ x, y, z ] t S , then x = 3 2 y - 5 2 z . Then x y z = 3 2 y - 5 2 z y z = y 3 2 1 0 + z

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ma265 lecture notes - Chapter 3 SUBSPACES 3.1 Introduction...

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