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Problem 3 (Basis for the kernel of a matrix) Let A be an mxn matrix and consider W = { € R" | Añ = 0}. We know from assignment 1 that W is a subspace of R". In this problem, we wish to explain why the fol- lowing strategy to find a basis of W works: if the solutions to A = 0 are given by = t₁₁ + ... + tkʊk, where t₁,..., tk are free parameters, then the vectors 1,...,k form a basis of W. (a) Show that if the matrix A' is obtained from the matrix A by a row operation (i.e. adding a multiple of a row to another row, multiplying a row by a nonzero constant or swapping two rows), then { € R³ | Añ = 0} = {ã € R³ | A'ñ = 0}. Deduce that if B is the reduced row echelon form of A, then {ã € R” | Añ = 0} = {ã € R³ | Bã = 0}. (b) Looking at the reduced row echelon form of A, we can see which variables are "leading", i.e. have a leading 1 in their corresponding column, and which variables are "free", i.e. have no leading 1 in their corresponding column. When solving the system B = 0, we see that each leading variable can be expressed in terms of the free variables that succeed it. In other words, if x is a leading variable, then we have x₁ = n Σcijxj j=i+1 I; free 1 for some constants cij E R. Let i₁ < < ik denote the indices of the free variables. Since the free variables xi, really are "free" to have any value t;, we conclude that the solutions to the system Bi are given by = = x1 : Ti-1 Til =t1 ++tk Tix-1 Tik Tik+1 In Jk Prove that the vectors ₁,..., Uk are linearly independent. (c) Conclude that {1,..., k} is a basis of W. Remark: To help digest all the notational intricacies involved in this problem, it might be helpful to work in parallel with the particular matrix [1 2 3 4 A = 0 0 1 2 00 24

Problem 3 (Basis for the kernel of a matrix) Let A be an mxn matrix and consider W = { € R" | Añ = 0}. We know from assignment 1 that W is a subspace of R". In this problem, we wish to explain why the fol- lowing strategy to find a basis of W works: if the solutions to A = 0 are given by = t₁₁ + ... + tkʊk, where t₁,..., tk are free parameters, then the vectors 1,...,k form a basis of W. (a) Show that if the matrix A' is obtained from the matrix A by a row operation (i.e. adding a multiple of a row to another row, multiplying a row by a nonzero constant or swapping two rows), then { € R³ | Añ = 0} = {ã € R³ | A'ñ = 0}. Deduce that if B is the reduced row echelon form of A, then {ã € R” | Añ = 0} = {ã € R³ | Bã = 0}. (b) Looking at the reduced row echelon form of A, we can see which variables are "leading", i.e. have a leading 1 in their corresponding column, and which variables are "free", i.e. have no leading 1 in their corresponding column. When solving the system B = 0, we see that each leading variable can be expressed in terms of the free variables that succeed it. In other words, if x is a leading variable, then we have x₁ = n Σcijxj j=i+1 I; free 1 for some constants cij E R. Let i₁ < < ik denote the indices of the free variables. Since the free variables xi, really are "free" to have any value t;, we conclude that the solutions to the system Bi are given by = = x1 : Ti-1 Til =t1 ++tk Tix-1 Tik Tik+1 In Jk Prove that the vectors ₁,..., Uk are linearly independent. (c) Conclude that {1,..., k} is a basis of W. Remark: To help digest all the notational intricacies involved in this problem, it might be helpful to work in parallel with the particular matrix [1 2 3 4 A = 0 0 1 2 00 24

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter4: Eigenvalues And Eigenvectors
Section4.6: Applications And The Perron-frobenius Theorem
Problem 24EQ
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Problem 3 (Basis for the kernel of a matrix) Let A be an mxn matrix and consider W = { € R" | Añ = 0}. We know from assignment 1 that W is a subspace of R". In this problem, we wish to explain why the fol- lowing strategy to find a basis of W works: if the solutions to A = 0 are given by = t₁₁ + ... + tkʊk, where t₁,..., tk are free parameters, then the vectors 1,...,k form a basis of W. (a) Show that if the matrix A' is obtained from the matrix A by a row operation (i.e. adding a multiple of a row to another row, multiplying a row by a nonzero constant or swapping two rows), then { € R³ | Añ = 0} = {ã € R³ | A'ñ = 0}. Deduce that if B is the reduced row echelon form of A, then {ã € R” | Añ = 0} = {ã € R³ | Bã = 0}. (b) Looking at the reduced row echelon form of A, we can see which variables are "leading", i.e. have a leading 1 in their corresponding column, and which variables are "free", i.e. have no leading 1 in their corresponding column. When solving the system B = 0, we see that each leading variable can be expressed in terms of the free variables that succeed it. In other words, if x is a leading variable, then we have x₁ = n Σcijxj j=i+1 I; free 1 for some constants cij E R. Let i₁ < < ik denote the indices of the free variables. Since the free variables xi, really are "free" to have any value t;, we conclude that the solutions to the system Bi are given by = = x1 : Ti-1 Til =t1 ++tk Tix-1 Tik Tik+1 In Jk Prove that the vectors ₁,..., Uk are linearly independent. (c) Conclude that {1,..., k} is a basis of W. Remark: To help digest all the notational intricacies involved in this problem, it might be helpful to work in parallel with the particular matrix [1 2 3 4 A = 0 0 1 2 00 24
Transcribed Image Text:Problem 3 (Basis for the kernel of a matrix) Let A be an mxn matrix and consider W = { € R" | Añ = 0}. We know from assignment 1 that W is a subspace of R". In this problem, we wish to explain why the fol- lowing strategy to find a basis of W works: if the solutions to A = 0 are given by = t₁₁ + ... + tkʊk, where t₁,..., tk are free parameters, then the vectors 1,...,k form a basis of W. (a) Show that if the matrix A' is obtained from the matrix A by a row operation (i.e. adding a multiple of a row to another row, multiplying a row by a nonzero constant or swapping two rows), then { € R³ | Añ = 0} = {ã € R³ | A'ñ = 0}. Deduce that if B is the reduced row echelon form of A, then {ã € R” | Añ = 0} = {ã € R³ | Bã = 0}. (b) Looking at the reduced row echelon form of A, we can see which variables are "leading", i.e. have a leading 1 in their corresponding column, and which variables are "free", i.e. have no leading 1 in their corresponding column. When solving the system B = 0, we see that each leading variable can be expressed in terms of the free variables that succeed it. In other words, if x is a leading variable, then we have x₁ = n Σcijxj j=i+1 I; free 1 for some constants cij E R. Let i₁ < < ik denote the indices of the free variables. Since the free variables xi, really are "free" to have any value t;, we conclude that the solutions to the system Bi are given by = = x1 : Ti-1 Til =t1 ++tk Tix-1 Tik Tik+1 In Jk Prove that the vectors ₁,..., Uk are linearly independent. (c) Conclude that {1,..., k} is a basis of W. Remark: To help digest all the notational intricacies involved in this problem, it might be helpful to work in parallel with the particular matrix [1 2 3 4 A = 0 0 1 2 00 24
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