Example: barber

Math 2331 { Linear Algebra

The Dimension of a Vector SpaceMath 2331 Linear The Dimension of a Vector SpaceJiwen HeDepartment of Mathematics, University of jiwenhe/math2331 Jiwen He, University of HoustonMath 2331, Linear Algebra1 / The Dimension of a Vector SpaceDimension Basis The Dimension of a Vector SpaceThe Dimension of a Vector Space: TheoremsThe Dimension of a Vector Space: DefinitionThe Dimension of a Vector Space: ExampleDimensions of Subspaces ofR3 Dimensions of Subspaces: TheoremThe Basis TheoremDimensions of ColAand NulA: ExamplesJiwen He, University of HoustonMath 2331, Linear Algebra2 / The Dimension of a Vector SpaceDimension Basis TheoremThe Dimension of a Vector Space: TheoremsTheorem (9)If a vector spaceVhas a basis ={b1,..,bn}, then any set inVcontaining more thannvectors must be linearly :Suppose{u1.}

3 is a linear combination of v 1 and v 2, so by the Spanning Set Theorem, we may discard v 3. v 4 is not a linear combination of v 1 and v 2. So fv 1;v 2;v 4gis a basis for W. Also, dim W = . Jiwen He, University of Houston Math 2331, Linear Algebra 7 / 14

Tags:

  Linear, Algebra, Linear algebra

Information

Domain:

Source:

Link to this page:

Please notify us if you found a problem with this document:

Other abuse

Transcription of Math 2331 { Linear Algebra

1 The Dimension of a Vector SpaceMath 2331 Linear The Dimension of a Vector SpaceJiwen HeDepartment of Mathematics, University of jiwenhe/math2331 Jiwen He, University of HoustonMath 2331, Linear Algebra1 / The Dimension of a Vector SpaceDimension Basis The Dimension of a Vector SpaceThe Dimension of a Vector Space: TheoremsThe Dimension of a Vector Space: DefinitionThe Dimension of a Vector Space: ExampleDimensions of Subspaces ofR3 Dimensions of Subspaces: TheoremThe Basis TheoremDimensions of ColAand NulA: ExamplesJiwen He, University of HoustonMath 2331, Linear Algebra2 / The Dimension of a Vector SpaceDimension Basis TheoremThe Dimension of a Vector Space: TheoremsTheorem (9)If a vector spaceVhas a basis ={b1,..,bn}, then any set inVcontaining more thannvectors must be linearly :Suppose{u1.}

2 ,up}is a set of vectors inVwherep> the coordinate vectors{[u1] , ,[up] }are >n,{[u1] , ,[up] }are linearly dependent andtherefore{u1,..,up}are linearly dependent. Jiwen He, University of HoustonMath 2331, Linear Algebra3 / The Dimension of a Vector SpaceDimension Basis TheoremThe Dimension of a Vector Space: Theorems (cont.)Theorem (10)If a vector spaceVhas a basis ofnvectors, then every basis ofVmust consist :Suppose 1is a basis forVconsisting of suppose 2is any other basis forV. By the definition of abasis, we know that 1and 2are both linearly independent Theorem 9, if 1has more vectors than 2, thenis alinearly dependent set (which cannot be the case).Again by Theorem 9, if 2has more vectors than 1, thenisa linearly dependent set (which cannot be the case).Therefore 2has exactly n vectors also.

3 Jiwen He, University of HoustonMath 2331, Linear Algebra4 / The Dimension of a Vector SpaceDimension Basis TheoremThe Dimension of a Vector Space: DefinitionDimension of a Vector SpaceIfVis spanned by a finite set, thenVis said to befinite-dimensional, and thedimensionofV, written as dimV, isthe number of vectors in a basis forV. The dimension of the zerovector space{0}is defined to be 0. IfVis not spanned by a finiteset, thenVis said to standard basis forP3is{}. So dimP3=.In general, dimPn=n+ standard basis forRnis{e1,..,en}wheree1,..,enare thecolumns , for example, dimR3= He, University of HoustonMath 2331, Linear Algebra5 / The Dimension of a Vector SpaceDimension Basis TheoremThe Dimension of a Vector Space: ExampleExampleFind a basis and the dimension of the subspaceW= a+b+ 2c2a+ 2b+ 4c+db+c+d3a+ 3c+d :a,b,c,dare real.

4 Solution:Since a+b+ 2c2a+2b+4c+db+c+d3a+ 3c+d =a 1203 +b 1210 +c 2413 +d 0111 Jiwen He, University of HoustonMath 2331, Linear Algebra6 / The Dimension of a Vector SpaceDimension Basis TheoremThe Dimension of a Vector Space: Example (cont.)W=span{v1,v2,v3,v4}wherev1= 1203 ,v2= 1210 ,v3= 2413 ,v4= 0111 .Note thatv3is a Linear combination ofv1andv2, so by theSpanning Set Theorem, we may not a Linear combination ofv1andv2. So{v1,v2,v4}isa basis forW. Also, dimW=.Jiwen He, University of HoustonMath 2331, Linear Algebra7 / The Dimension of a Vector SpaceDimension Basis TheoremDimensions of Subspaces ofR3 Example (Dimensions of subspaces ofR3)10-dimensional subspacecontains only the zero vector0= (0,0,0).21-dimensional {v}wherev6=0is subspaces arethrough the {u,v}whereuandvare inR3and are not multiples of each subspaces arethrough the {u,v,w}whereu, v,warelinearly independent vectors inR3.

5 This subspace isR3itselfbecause the columns ofA= [u v w] spanR3according tothe He, University of HoustonMath 2331, Linear Algebra8 / The Dimension of a Vector SpaceDimension Basis TheoremDimensions of Subspaces: TheoremTheorem (11)LetHbe a subspace of a finite-dimensional vector spaceV. Anylinearly independent set inHcan be expanded, if necessary, to abasis forH. Also,His finite-dimensional anddimH 100 , 110 . ThenHis a subspace ofR3anddimH<dimR3. We could expand the spanning set 100 , 110 to 100 , 110 , 001 for a basis He, University of HoustonMath 2331, Linear Algebra9 / The Dimension of a Vector SpaceDimension Basis TheoremThe Basis TheoremTheorem (12 The Basis Theorem)LetVbe ap dimensional vector space,p 1. Any linearlyindependent set of exactlypvectors inVis automatically a basisforV.

6 Any set of exactlypvectors that spansVis automaticallya basis that{t,1 t,1 +t t2}is a basis :Letv1=t,v2= 1 t,v3= 1 +t t2and ={1,t,t2}.Corresponding coordinate vectors[v1] = 010 ,[v2] = 1 10 ,[v3] = 11 1 Jiwen He, University of HoustonMath 2331, Linear Algebra10 / The Dimension of a Vector SpaceDimension Basis TheoremThe Basis Theorem (cont.)[v2] is not a multiple of [v1] [v3] is not a Linear combination of [v1] and [v2] = {[v1] ,[v2] ,[v3] }is linearly independent and therefore{v1,v2,v3}is also linearly dimP2= 3,{v1,v2,v3}is a basis forP2according to TheBasis He, University of HoustonMath 2331, Linear Algebra11 / The Dimension of a Vector SpaceDimension Basis TheoremDimensions of ColAand NulA: ExampleRecall our techniques to find basis sets for column spaces and [1 2 3 42 4 7 8].

7 Find dim ColAand dim [1 2 3 42 4 7 8] [1 2 3 40 0 1 0]So{[ ],[ ]}is a basis for ColAand dim ColA= He, University of HoustonMath 2331, Linear Algebra12 / The Dimension of a Vector SpaceDimension Basis TheoremDimensions of ColAand NulA: Example (cont.)Now solveAx=0by row-reducing the corresponding augmentedmatrix. Then we arrive at[1 2 3 4 02 4 7 8 0] [1 2 0 4 00 0 1 0 0]x1= 2x2 4x4x3= 0 x1x2x3x4 =x2 2100 +x4 4001 Jiwen He, University of HoustonMath 2331, Linear Algebra13 / The Dimension of a Vector SpaceDimension Basis TheoremDimensions of ColAand NulA: Example (cont.)So 2100 , 4001 is a basis for NulAand dim NulA= ColA=number of pivot columns ofAdim NulA=number of free variables He, University of HoustonMath 2331, Linear Algebra14 / 14


Related search queries