1) Consider the real vector space V = R", and let y € V*. The point of this problem isto get another interpretation of the dual space in this case.we can "think" of y as a "row" vector by considering its matrixrepresentation under the standard basis (it will be a 1 × n matrix, ie a row vector.) Let usnow prove that every row vector gives rise to a linear functional. Let (x₁,...,xn) be a "row"vector (ie, a 1 × n matrix) with each x; € R and define the functional : V → R by00) := (x₁,...,xn) (a1a2:Ana1a2an= x₁α1 + x₂a2 + . Xn AnFirst off, convince yourself that this is actually a linear functional(a)Recall that there is a bijection V* = L(V, R) ~ Mixn (R) that sends a linearmap to its matrix representation. Show that this construction above (that sends a 1xnmatrix to the linear functional p) is just the inverse of this isomorphism. (In otherwords, if you start with a linear functional, take its matrix, and then define this newlinear functional as above, you get the original linear functional you started with)(b)Under this identification above, and letting V = R³ what is the row vectorthat corresponds to the dual basis vector ef? How about e?

Let us start solving the problem step by step

1) Consider the real vector space V = R¹, and let y € V*. The point of this problem is to get another interpretation of the dual space in this case. we can "think" of y as a "row" vector by considering its matrix representation under the standard basis (it will be a 1 × n matrix, ie a row vector.) Let us now prove that every row vector gives rise to a linear functional. Let (x₁,...,xn) be a “row” vector (ie, a 1 x n matrix) with each x; € R and define the functional y : V → R by a1 - () -- a2 an := (x₁,...,xn) ( a1 a2 an = x₁α1 + x₂A2+...XnAn First off, convince yourself that this is actually a linear tional (a) ~ Recall that there is a bijection V* = L(V,R) ≈ M₁×n(R) that sends a linear map to its matrix representation. Show that this construction above (that sends a 1×n matrix to the linear functional y) is just the inverse of this isomorphism. (In other words, if you start with a linear functional, take its matrix, and then define this new linear functional as above, you get the original linear functional you started with) (b) Under this identification above, and letting V = R³ what is the row vector that corresponds to the dual basis vector ef? How about ež?

1) Consider the real vector space V = R¹, and let y € V. The point of this problem is to get another interpretation of the dual space in this case. we can "think" of y as a "row" vector by considering its matrix representation under the standard basis (it will be a 1 × n matrix, ie a row vector.) Let us now prove that every row vector gives rise to a linear functional. Let (x₁,...,xn) be a “row” vector (ie, a 1 x n matrix) with each x; € R and define the functional y : V → R by a1 - () -- a2 an := (x₁,...,xn) ( a1 a2 an = x₁α1 + x₂A2+...XnAn First off, convince yourself that this is actually a linear tional (a) ~ Recall that there is a bijection V = L(V,R) ≈ M₁×n(R) that sends a linear map to its matrix representation. Show that this construction above (that sends a 1×n matrix to the linear functional y) is just the inverse of this isomorphism. (In other words, if you start with a linear functional, take its matrix, and then define this new linear functional as above, you get the original linear functional you started with) (b) Under this identification above, and letting V = R³ what is the row vector that corresponds to the dual basis vector ef? How about ež?

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter6: Vector Spaces

Section6.2: Linear Independence, Basis, And Dimension

Problem 15EQ

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