Problem 3 (Basis for the kernel of a matrix) Let A be an mxn matrix and consider W = { € R" | A= 0}. We know from assignment 1 that W is a subspace of Rn. In this problem, we wish to explain why the fol- lowing strategy to find a basis of W works: if the solutions to Añ = 0 are given by = t₁₁++ tkk, where t₁t are free parameters, then the vectors ₁, 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 {7 € R" | Az = 0} = {7 € R” | A'z = 0}. Deduce that if B is the reduced row echelon form of A, then { € R" | Az=0} = {ĩ € R" | 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 Bã = 0, we see that each leading variable can be expressed in terms of the free variables that succeed it. In other words, if a; is a leading variable, then we have x₁ = n j=i+1 x; free Cijxj 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 Bã = 0 are given by x1 * * 1 Xi1 Tin+1 * * 0 x = = t1 ++tk Xik-1 0 1 Tik 0 Xik+1 xn Prove that the vectors ...., are linearly independent. Ük
Problem 3 (Basis for the kernel of a matrix) Let A be an mxn matrix and consider W = { € R" | A= 0}. We know from assignment 1 that W is a subspace of Rn. In this problem, we wish to explain why the fol- lowing strategy to find a basis of W works: if the solutions to Añ = 0 are given by = t₁₁++ tkk, where t₁t are free parameters, then the vectors ₁, 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 {7 € R" | Az = 0} = {7 € R” | A'z = 0}. Deduce that if B is the reduced row echelon form of A, then { € R" | Az=0} = {ĩ € R" | 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 Bã = 0, we see that each leading variable can be expressed in terms of the free variables that succeed it. In other words, if a; is a leading variable, then we have x₁ = n j=i+1 x; free Cijxj 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 Bã = 0 are given by x1 * * 1 Xi1 Tin+1 * * 0 x = = t1 ++tk Xik-1 0 1 Tik 0 Xik+1 xn Prove that the vectors ...., are linearly independent. Ük
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Problem 3 (Basis for the kernel of a matrix) Let A be an mxn matrix and consider W = { € R" | A=
0}. We know from assignment 1 that W is a subspace of Rn. In this problem, we wish to explain why the fol-
lowing strategy to find a basis of W works: if the solutions to Añ = 0 are given by = t₁₁++ tkk, where
t₁t are free parameters, then the vectors ₁, 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
{7 € R" | Az = 0} = {7 € R” | A'z = 0}.
Deduce that if B is the reduced row echelon form of A, then
{ € R" | Az=0} = {ĩ € R" | 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 Bã = 0, we see that each leading variable can be expressed
in terms of the free variables that succeed it. In other words, if a; is a leading variable, then we have
x₁ =
n
j=i+1
x; free
Cijxj
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 Bã = 0
are given by
x1
*
*
1
Xi1
Tin+1
*
*
0
x =
= t1
++tk
Xik-1
0
1
Tik
0
Xik+1
xn
Prove that the vectors ...., are linearly independent.
Ük
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