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label part tiny 1 of 1 Exercise 1. Let A E Kn over K. The column space of A is the subspace of Km spanned by the columns of A, in short Col(A is the ith column of A, similarly for the rouw space Row(A). The null space of A is Nul(A) E K" | Ar = 0}C K" For this problem, K = Z2. Let f : K5 A) where A Span(A K4 by f() Ar for E K where 1 1 1 1 1 1 A= 1 0 0 0 1 EKXS 1 1 0 0 1 1 0 1 (c.1 Apply row operations to A to get a reduced row Echelon form A (c.2) Use Problem (b) in HW2 to claim that Row(A Row(A'), then get a basis for the row space Row(A) HW2 b is provided at the end of the question (c.3) Find a basis for Nul(A) Nul (A') (you should know that the two spaces are the same) (c.4 Using the basis in (c.3) to express each column of A a combination of some independent columns of A, hence get a basis for the column space (A) (c.5) Is finjective or onto? What's the dimension for the kernel space and image space of f? (c.6) (About dual space) Note that each vector vE K5 defines a linear functional Ks K by v*(r (v, a) - ΣΣ 0π, ε κ . (A) and define W= {v* € (K5)* : v* (u) = 0 for all u E W}. Get a for K Let basis for W using (c.3) For your reference, this was HW2, Problem b Let E be any vector space over K and u;}ieI any family of vectors in E (the family could be infinite). Let us, be any two vectors in the family (st) and let a E K be arbitrary. Show that Span(u, u}iel,) = Span(u, + aut, {u;}ie!,) where I\I{s}. This means that one can add any multiple of any vector to any other vector in the family, and the spanning space does not change. Hence deduce that Span(w, fu}ieh) = Span(u, {u}ien) where w is a finite combination of the vectors in the family that contains u, with a nonzero coefficient. (Here one should make connection to row operations of a matrix and how the row space changes.)

label part tiny 1 of 1 Exercise 1. Let A E Kn over K. The column space of A is the subspace of Km spanned by the columns of A, in short Col(A is the ith column of A, similarly for the rouw space Row(A). The null space of A is Nul(A) E K" | Ar = 0}C K" For this problem, K = Z2. Let f : K5 A) where A Span(A K4 by f() Ar for E K where 1 1 1 1 1 1 A= 1 0 0 0 1 EKXS 1 1 0 0 1 1 0 1 (c.1 Apply row operations to A to get a reduced row Echelon form A (c.2) Use Problem (b) in HW2 to claim that Row(A Row(A'), then get a basis for the row space Row(A) HW2 b is provided at the end of the question (c.3) Find a basis for Nul(A) Nul (A') (you should know that the two spaces are the same) (c.4 Using the basis in (c.3) to express each column of A a combination of some independent columns of A, hence get a basis for the column space (A) (c.5) Is finjective or onto? What's the dimension for the kernel space and image space of f? (c.6) (About dual space) Note that each vector vE K5 defines a linear functional Ks K by v*(r (v, a) - ΣΣ 0π, ε κ . (A) and define W= {v* € (K5)* : v* (u) = 0 for all u E W}. Get a for K Let basis for W using (c.3) For your reference, this was HW2, Problem b Let E be any vector space over K and u;}ieI any family of vectors in E (the family could be infinite). Let us, be any two vectors in the family (st) and let a E K be arbitrary. Show that Span(u, u}iel,) = Span(u, + aut, {u;}ie!,) where I\I{s}. This means that one can add any multiple of any vector to any other vector in the family, and the spanning space does not change. Hence deduce that Span(w, fu}ieh) = Span(u, {u}ien) where w is a finite combination of the vectors in the family that contains u, with a nonzero coefficient. (Here one should make connection to row operations of a matrix and how the row space changes.)

Elementary Linear Algebra (MindTap Course List)
Elementary Linear Algebra (MindTap Course List)
8th Edition
ISBN:9781305658004
Author:Ron Larson
Publisher:Ron Larson
Chapter5: Inner Product Spaces
Section5.CR: Review Exercises
Problem 41CR: Let B={(0,2,2),(1,0,2)} be a basis for a subspace of R3, and consider x=(1,4,2), a vector in the...
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label part tiny 1 of 1 Exercise 1. Let A E Kn over K. The column space of A is the subspace of Km spanned by the columns of A, in short Col(A is the ith column of A, similarly for the rouw space Row(A). The null space of A is Nul(A) E K" | Ar = 0}C K" For this problem, K = Z2. Let f : K5 A) where A Span(A K4 by f() Ar for E K where 1 1 1 1 1 1 A= 1 0 0 0 1 EKXS 1 1 0 0 1 1 0 1 (c.1 Apply row operations to A to get a reduced row Echelon form A (c.2) Use Problem (b) in HW2 to claim that Row(A Row(A'), then get a basis for the row space Row(A) HW2 b is provided at the end of the question (c.3) Find a basis for Nul(A) Nul (A') (you should know that the two spaces are the same) (c.4 Using the basis in (c.3) to express each column of A a combination of some independent columns of A, hence get a basis for the column space (A) (c.5) Is finjective or onto? What's the dimension for the kernel space and image space of f? (c.6) (About dual space) Note that each vector vE K5 defines a linear functional Ks K by v*(r (v, a) - ΣΣ 0π, ε κ . (A) and define W= {v* € (K5)* : v* (u) = 0 for all u E W}. Get a for K Let basis for W using (c.3) For your reference, this was HW2, Problem b Let E be any vector space over K and u;}ieI any family of vectors in E (the family could be infinite). Let us, be any two vectors in the family (st) and let a E K be arbitrary. Show that Span(u, u}iel,) = Span(u, + aut, {u;}ie!,) where I\I{s}. This means that one can add any multiple of any vector to any other vector in the family, and the spanning space does not change. Hence deduce that Span(w, fu}ieh) = Span(u, {u}ien) where w is a finite combination of the vectors in the family that contains u, with a nonzero coefficient. (Here one should make connection to row operations of a matrix and how the row space changes.)
Transcribed Image Text:label part tiny 1 of 1 Exercise 1. Let A E Kn over K. The column space of A is the subspace of Km spanned by the columns of A, in short Col(A is the ith column of A, similarly for the rouw space Row(A). The null space of A is Nul(A) E K" | Ar = 0}C K" For this problem, K = Z2. Let f : K5 A) where A Span(A K4 by f() Ar for E K where 1 1 1 1 1 1 A= 1 0 0 0 1 EKXS 1 1 0 0 1 1 0 1 (c.1 Apply row operations to A to get a reduced row Echelon form A (c.2) Use Problem (b) in HW2 to claim that Row(A Row(A'), then get a basis for the row space Row(A) HW2 b is provided at the end of the question (c.3) Find a basis for Nul(A) Nul (A') (you should know that the two spaces are the same) (c.4 Using the basis in (c.3) to express each column of A a combination of some independent columns of A, hence get a basis for the column space (A) (c.5) Is finjective or onto? What's the dimension for the kernel space and image space of f? (c.6) (About dual space) Note that each vector vE K5 defines a linear functional Ks K by v*(r (v, a) - ΣΣ 0π, ε κ . (A) and define W= {v* € (K5)* : v* (u) = 0 for all u E W}. Get a for K Let basis for W using (c.3) For your reference, this was HW2, Problem b Let E be any vector space over K and u;}ieI any family of vectors in E (the family could be infinite). Let us, be any two vectors in the family (st) and let a E K be arbitrary. Show that Span(u, u}iel,) = Span(u, + aut, {u;}ie!,) where I\I{s}. This means that one can add any multiple of any vector to any other vector in the family, and the spanning space does not change. Hence deduce that Span(w, fu}ieh) = Span(u, {u}ien) where w is a finite combination of the vectors in the family that contains u, with a nonzero coefficient. (Here one should make connection to row operations of a matrix and how the row space changes.)
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