2010Fall-LA-Lecture-13[1]

Linear Algebra Lecture 13 Linear Independence, Spans, and Bases Linear Independence A set of vectors 𝑺 = {𝒗𝟏 , 𝒗𝟐 , ⋯ , 𝒗𝒏 } is linearly independent if 𝜶𝟏 𝒗𝟏 + 𝜶𝟐 𝒗𝟐 + ⋯ + 𝜶𝒏 𝒗𝒏 = 𝟎 ⟹ 𝜶𝟏 = 𝟎, 𝜶𝟐 = 𝟎, ⋯ , 𝜶𝒏 = 𝟎 Otherwise it is linearly dependent. 𝑺 is linearly dependent means that there are vectors in 𝑺 which can be represented as linear combination of the other vectors 𝜶𝟏 𝒗𝟏 + 𝜶𝟐 𝒗𝟐 + ⋯ + 𝜶𝒏 𝒗𝒏 = 𝟎 𝜶𝟏 ≠ 𝟎 𝜶𝟏 𝒗𝟏 = −𝜶𝟐 𝒗𝟐 − ⋯ − 𝜶𝒏 𝒗𝒏 𝒗𝟏 = −𝜶𝟐 𝒗 𝜶𝟏 𝟐 − ⋯− 𝜶𝒏 𝒗 𝜶𝟏 𝒏 For example: 𝟏 𝟏 𝟏 [ 𝟐] [ 𝟒] [ 𝟎] 𝟑 𝟔 𝟔 are linearly independent Can we find a,b,c such that 𝟏 𝟏 𝟏 𝟎 𝒂 [𝟐] + 𝒃 [𝟒] + 𝒄 [𝟎] = [𝟎] 𝟑 𝟔 𝟔 𝟎 Notice 𝟏 𝟏 𝟏 𝒂 𝟎 [𝟐 𝟒 𝟎] [𝒃] = [𝟎] 𝟑 𝟔 𝟔 𝒄 𝟎 𝟏 𝟏 𝟏 Det([𝟐 𝟒 𝟎]) = 𝟏𝟐 ≠ 𝟎 𝟑 𝟔 𝟔 The system has only one solution and that one is a=0,b=0,c=0. For example: 𝟏 [ 𝟐] 𝟑 𝟏 [ 𝟒] 𝟔 𝟏 [ 𝟎] 𝟎 are linearly dependent(Keary) Can we find a,b,c such that 𝟏 𝟏 𝟏 𝟎 𝒂 [𝟐] + 𝒃 [𝟒] + 𝒄 [𝟎] = [𝟎] 𝟑 𝟔 𝟎 𝟎 Solution: a=-2, b=1,c=1 𝟏 𝟏 𝟏 𝟏 𝟏 [ ] + [𝟎] = [𝟐] 𝟒 𝟐 𝟐 𝟔 𝟎 𝟑 Yes. This set of vectors is linearly dependent. The minimal number of independent vectors in the subspaces is called the rank or dimension of the subspace. How do we find the dimensions of the four spaces? Answer: Gauss-Jordan Elimination process 𝟏 𝟑 𝟐 𝟐 𝑨 = [ 𝟎 𝟏 𝟐 𝟑] 𝟎 𝟎 𝟏 𝟐 The first 3 columns are independent. Why? 𝟏 𝟑 [𝟎 𝟏 𝟎 𝟎 𝟐 𝒂 𝟎 𝟐] [𝒃] = [𝟎] 𝟏 𝒄 𝟎 𝒅𝒆𝒕 = 𝟏 ≠ 𝟎 This means we have only one, the trivial solution. 𝒂 𝟏 𝟑 𝟐 𝟐 𝒃 𝟎 [𝟎 𝟏 𝟐 𝟑] [ ] = [𝟎] 𝒄 𝟎 𝟎 𝟏 𝟐 𝒅 𝟎 𝒂 −𝟏 𝒃 𝟏 [ ]=[ ] 𝒄 −𝟐 𝒅 𝟏 𝟏 𝟎 𝟎 𝟑 𝟏 𝟎 [ ] 𝟐 𝟐 𝟏 𝟐 𝟑 𝟐 Because the system has only one solution 𝒂 𝟎 [ 𝒃 ] = [ 𝟎] 𝒄 𝟎 The fourth one is linearly dependent of the others. How about rows? 𝟏 𝟑 𝟐 𝟐 𝑨 = [ 𝟎 𝟏 𝟐 𝟑] 𝟎 𝟎 𝟏 𝟐 if 𝒂𝟏 𝑹𝟏 + 𝒂𝟐 𝑹𝟐 + 𝒂𝟑 𝑹𝟑 = 𝟎 then ... What else could happen? 𝟏 𝟑 𝟐 𝟐 𝑨 = [ 𝟎 𝟏 𝟐 𝟑] 𝟎 𝟎 𝟎 𝟎 The Rank is 2. 𝟏 𝟑 𝟐 𝟐 𝑨 = [ 𝟎 𝟎 𝟐 𝟑] 𝟎 𝟎 𝟎 𝟎 The Rank is 2. 𝟏 𝟑 𝟐 𝟐 𝑨 = [ 𝟎 𝟎 𝟎 𝟎] 𝟎 𝟎 𝟎 𝟎 The Rank is 1. Notice: Rank(Col(A))=Rank(Row(A)) _____________________________________ Let 𝟏 𝟑 𝟐 𝑨 = [𝟎 𝟏 𝟐 𝟎 𝟏 𝟏 𝟏 𝟑 𝟐 𝑨 = [𝟎 𝟏 𝟐 𝟎 𝟎 −𝟏 𝟐 𝟑] 𝟏 𝟐 𝟑] −𝟐 The rank of the Null(A) is called nullity. Find the Rank and Nullity Notice: Rank + Nullity=the number of Columns The Rank and Nullity Theorem More on Linear Independence ____________________________________________ (1)A finite set of vectors that contains a zero vector is linearly dependent 𝟏 𝟏 𝟏 𝟏 𝟎 𝟒 𝟐 𝟒 𝟐 𝟎 𝟎 𝟖 Let 𝑺 = 𝟑 , 𝟑 , 𝟎 , 𝟑 , 𝟎 , 𝟏𝟐 𝟓 𝟓 𝟓 𝟓 𝟎 𝟐𝟎 {[𝟒] [𝟒] [𝟒] [𝟒] [𝟎] [𝟏𝟐]} Is S linearly independent or dependent? Dependent. ____________________________________________ (2) A set of two vectors is linearly independent iff neither of the vectors is a multiple of the other 𝟏 𝟒 𝟐 𝟖 Let 𝑺 = 𝟑 , 𝟏𝟐 𝟓 𝟐𝟎 {[𝟒] [𝟏𝟐]} Is S linearly independent or dependent? Independent, because the two vectors are not multiple of each other. __________________________________________ (2)Let 𝑺 = {𝒗𝟏 , 𝒗𝟐 , ⋯ , 𝒗𝒏 } be a set of vectors in 𝑹𝒎 . If 𝒏 > 𝑚, then S is linearly dependent. 𝟏 𝟏 𝟏 𝟏 𝟏 𝟒 𝟐 𝟒 𝟐 𝟎 𝟎 𝟖 𝑺 = 𝟑 , 𝟑 , 𝟎 , 𝟑 , 𝟏 , 𝟏𝟐 𝟓 𝟓 𝟓 𝟓 𝟎 𝟐𝟎 {[𝟒] [𝟒] [𝟒] [𝟒] [𝟎] [𝟏𝟐]} S is dependent. _________________________________________ Let 𝑺 = {𝒗𝟏 , 𝒗𝟐 , ⋯ , 𝒗𝒏 } be a set of vectors Then 𝒔𝒑𝒂𝒏(𝑺) = {𝒖|𝒖 = 𝒂𝟏 𝒗𝟏 + 𝒂𝟐 𝒗𝟐 + ⋯ + 𝒂𝒏 𝒗𝒏 } i.e. Span(S) is a set of all vectors which can be represented as a linear combination of vectors in S. Span(S) is a vector space. Example: Let 𝟏 𝟐 𝑺 = {[𝟐] , [𝟏]} 𝟑 𝟏 𝟏 𝟐 𝒔𝒑𝒂𝒏(𝑺) = {𝒂 [𝟐] + 𝒃 [𝟏] , 𝒂, 𝒃 𝒂𝒓𝒆 𝒔𝒄𝒂𝒍𝒂𝒓𝒔} 𝟑 𝟏 A Basis of the Vector space If V is a vector space and 𝑺 = {𝒗𝟏 , 𝒗𝟐 , ⋯ , 𝒗𝒏 } is a set of vectors in V. Then S is called a basis if (1) S is linearly independent (2) span(S)=V Example: canonical example, V=𝑹𝟑 Canonical Basis 𝟏 𝟎 𝟎 𝑺 = {[𝟎] , [𝟏] , [𝟎]} 𝟎 𝟎 𝟏 Pick𝒖 ∈ 𝑹𝟑 . −𝟓 John’s 𝒖 = [ 𝟖 ] −𝟐𝟕 𝟏 𝟎 𝟎 −𝟓 𝒖 = [ 𝟖 ] = −𝟓 [𝟎] + 𝟖 [𝟏] − 𝟐𝟕 [𝟎] 𝟎 𝟎 𝟏 −𝟐𝟕 Give me another basis for 𝑹𝟑 𝟏 𝟑 𝟏 𝑻 = {[𝟎] , [𝟏] , [𝟎]} 𝟎 𝟎 𝟏 Q:Is T a basis? Check independence! Det=1 hence they are linearly independent. −𝟓 John’s 𝒖 = [ 𝟖 ] −𝟐𝟕 𝟏 𝟑 𝟏 −𝟓 𝒖 = [ 𝟖 ] = 𝒂 [𝟎] + 𝒃 [𝟏] + 𝒄 [𝟎] = 𝟎 𝟎 𝟏 −𝟐𝟕 𝟏 𝟑 𝟏 𝒂 −𝟓 [𝟎 𝟏 𝟎] [𝒃] = [ 𝟖 ] 𝟎 𝟎 𝟏 𝒄 −𝟐𝟕 Does this system have a solution? Det=1 means it has the solution no matter what u you pick. T is a basis for 𝑹𝟑 . Theorem: A vector representation in a basis is unique. Proof: Let u have two representation in the same basis. 𝒖 = 𝒂𝟏 𝒗𝟏 + 𝒂𝟐 𝒗𝟐 + ⋯ + 𝒂𝒏 𝒗𝒏 and 𝒖 = 𝒃𝟏 𝒗𝟏 + 𝒃𝟐 𝒗𝟐 + ⋯ + 𝒃𝒏 𝒗𝒏 Then 𝒖 − 𝒖 = (𝒂𝟏 − 𝒃𝟏 )𝒗𝟏 + (𝒂𝟐 − 𝒃𝟐 )𝒗𝟐 + ⋯ + (𝒂𝒏 − 𝒃𝒏 )𝒗𝒏 𝟎 = (𝒂𝟏 − 𝒃𝟏 )𝒗𝟏 + (𝒂𝟐 − 𝒃𝟐 )𝒗𝟐 + ⋯ + (𝒂𝒏 − 𝒃𝒏 )𝒗𝒏 Since 𝒗𝒋 s are independent this can happen only if (𝒂𝒋 − 𝒃𝒋 ) = 𝟎 𝒋 = 𝟏, 𝟐, ⋯ , 𝒏 Therefore 𝒂𝒋 = 𝒃𝒋 and representation is unique. Theorem. Every vector space has a basis. All bases of the same vector space have the same number of elements. This number is called a DIMENSION of the vector space. Example and Problems: Let 𝟏 𝟐 𝑺 = {[𝟐] , [𝟏]} 𝟑 𝟏 𝟏 𝟐 𝒔𝒑𝒂𝒏(𝑺) = {𝒂 [𝟐] + 𝒃 [𝟏] , 𝒂, 𝒃 𝒂𝒓𝒆 𝒔𝒄𝒂𝒍𝒂𝒓𝒔} 𝟑 𝟏 𝟐 Q: Is [ −𝟖 ] in span(S)? −𝟏𝟐 Yes, because 𝟏 𝟐 𝟐 [ −𝟖 ] = 𝒂 [𝟐] + 𝒃 [𝟏] 𝟑 −𝟏𝟐 𝟏 𝟏 𝟐 𝒂 𝟐 [𝟐 𝟏] [ ] = [ −𝟖 ] 𝒃 𝟑 𝟏 −𝟏𝟐 .. 𝟏 𝟎 −𝟒 [𝟏 𝟏] [−𝟐] 𝟎 𝟎 𝟏 Chad says: It is not in the span!!!!! _____________________ 𝟏 𝟐 𝟐 𝟐 𝟏 𝟐 𝑺 = {[ ] , [ ] , [ ]} 𝟎 𝟏 𝟎 𝟑 𝟏 𝟏 Is this set linearly independent? 𝟏 𝟐 𝟐 𝟐 𝟏 𝟐 [ ] 𝟎 𝟏 𝟎 𝟑 𝟏 𝟏 𝟏 𝟎 𝟐 𝟐 −𝟑 𝟐 [ ] 𝟎 𝟏 𝟎 𝟑 −𝟓 𝟏 𝟏 𝟎 𝟎 𝟐 −𝟑 𝟎 [ ] 𝟎 𝟏 −𝟐/𝟑 𝟑 −𝟓 𝟓/𝟑 We performed Gauss –Jordan on columns and we learned the following: S is linearly independent. S is a basis for its span. Also 𝟎 𝟏 𝟎 𝟎 𝟐 −𝟑 𝑻 = {[ ] , [ ] , [ ]} −𝟐/𝟑 𝟎 𝟏 𝟓/𝟑 𝟑 −𝟓 T is also a basis for span(S). Example 2. 𝟏 𝟐 𝟔 𝟐 𝟏 𝟔 𝑺 = {[ ] , [ ] , [ ]} 𝟎 𝟏 𝟐 𝟑 𝟏 𝟖 𝟏 𝟐 𝟔 𝟐 𝟏 𝟔 [ ] 𝟎 𝟏 𝟐 𝟑 𝟏 𝟖 𝟏 𝟎 𝟎 𝟐 −𝟑 𝟎 [ ] 𝟎 𝟏 𝟎 𝟑 −𝟓 𝟎 We performed Gauss –Jordan on columns and we learned the following: S is linearly dependent. S isn’t basis for span(S). But 𝟏 𝟎 𝟐 −𝟑 𝑻 = {[ ] , [ ]} is a basis for span(S). 𝟎 𝟏 𝟑 −𝟓 Dimension span(S)=2 Example 3. Compression. 𝟏 𝟐 𝟐 𝟏 𝟎 𝟏 𝑺= , 𝟑 𝟏 𝟏 𝟎 {[𝟎] [𝟏]} 𝟕 𝟏𝟏 𝟏 𝟏𝟔 𝟓 [𝟏] S is linearly independent because they are not multiple of each other. S is a basis for its span. Assume that the signal vectors are coming from span(S). For example a signal vector from this span 𝟕 𝟏𝟏 𝟏 is . 𝟏𝟔 𝟓 [𝟏] 𝟕 𝟏 𝟐 𝟏𝟏 𝟐 𝟏 𝟏 𝟎 𝟏 =𝟓 +𝟏 𝟏𝟔 𝟑 𝟏 𝟓 𝟏 𝟎 [𝟏] [𝟏 ] [𝟎 ] I need to send the signal to the other side. 𝟕 𝟏𝟏 𝟏 I could send 𝒐𝒓 I could send what? 𝟏𝟔 𝟓 [𝟏] 𝟓 James says : Send instead [ ]. So we saved 𝟏 66%. Let us test the system. We are on the other side and we received 𝟐 [ ]. What was the original signal vector? 𝟑 𝟖 𝟏 𝟐 𝟕 𝟐 𝟏 𝟑 𝟎 𝟏 =𝟐 +𝟑 𝟗 𝟑 𝟏 𝟐 𝟏 𝟎 [𝟏] [𝟑 ] [𝟎 ] _______________________________________ Q: Find a different basis for span(S). 𝟐 Q:Find a representation of [ −𝟖 ] in the new −𝟏𝟐 basis. Big Question: Transition between two bases. Let 𝑺 = {𝒗𝟏 , 𝒗𝟐 } and 𝑻 = { 𝒇𝟏 , 𝒇𝟐 } be two linear bases for the same vector space X. Let 𝒖 ∈ 𝑿. Then 𝒖 = 𝒂𝒗𝟏 + 𝒃𝒗𝟐 and 𝒖 = 𝒙𝒇𝟏 + 𝒚𝒇𝟐 Assume you know the first representation. How do you find the second representation? Or If you know a and b how can you find x and y(Really fast)? First notice that 𝒗𝟏 , 𝒗𝟐 ∈ 𝑿 Therefore it can be written in terms of the basis T. 𝒗𝟏 = 𝒑𝟏 𝒇𝟏 + 𝒑𝟐 𝒇𝟐 𝒗𝟐 = 𝒒𝟏 𝒇𝟏 + 𝒒𝟐 𝒇𝟐 𝒖 = 𝒂𝒗𝟏 + 𝒃𝒗𝟐 = = 𝒂(𝒑𝟏 𝒇𝟏 + 𝒑𝟐 𝒇𝟐 ) + 𝒃(𝒒𝟏 𝒇𝟏 + 𝒒𝟐 𝒇𝟐 ) = = (𝒂𝒑𝟏 + 𝒃𝒒𝟏 )𝒇𝟏 + (𝒂𝒑𝟐 + 𝒃𝒒𝟐 )𝒇𝟐 = = 𝒙𝒇𝟏 + 𝒚𝒇𝟐 Hence by __uniqueness____ 𝒙 = 𝒂𝒑𝟏 + 𝒃𝒒𝟏 𝒚 = 𝒂𝒑𝟐 + 𝒃𝒒𝟐 We can write this in matrix notation: 𝒑𝟏 𝒒𝟏 𝒂 𝒙 [𝒑 ] [ ] = [𝒚] 𝒒 𝟐 𝟐 𝒃 𝒑𝟏 𝒒𝟏 𝑻𝒗→𝒇 = [𝒑 ] transition matrix 𝒒 𝟐 𝟐 𝑻𝒗→𝒇 𝒖𝒗 = 𝒖𝒇 𝒖𝒗 = 𝑻𝒗→𝒇 −𝟏 𝒖𝒇 In the case 2x2 transformation matrix 𝑻𝒗→𝒇 −𝟏 𝒒𝟐 −𝒒𝟏 = [−𝒑 ] 𝒑 𝒑𝟏 𝒒𝟐 − 𝒑𝟐 𝒒𝟏 𝟐 𝟏 𝟏

2010Fall-LA-Lecture-13[1]

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