Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 25E from Chapter 6.4 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 25E
Chapter:
Problem:
0
Step-by-Step Solution
Given Information
We are given with a matrix: \[A = \left[ \begin{array} { r r r r } { - 10 } & { 13 } & { 7 } & { - 11 } \\ { 2 } & { 1 } & { - 5 } & { 3 } \\ { - 6 } & { 3 } & { 13 } & { - 3 } \\ { 16 } & { - 16 } & { - 2 } & { 5 } \\ { 2 } & { 1 } & { - 5 } & { - 7 } \end{array} \right]\]We have to find QR factorization for the above matrix:
Step-1:
Enter the matrix A in MATLAB.
>> A=[-10 13 7 -11; 2 1 5 3; -6 3 13 -3; 16 -16 -2 5; 2 1 -5 -7];
The required function:
function [B] = GramSchmidt_N(A)
[m,n] = size(A);
[U, jb] = rref(A);
x = length(jb);
B = zeros(m,x);
for i = 1:x
C(:,i)= A(:,(jb(i)));
end
B=C;
for i = 2:x
for j = 1:i-1
B(:,i) = B(:,i) - dot(C(:,i),B(:,j))/dot(B(:,j),B(:,j))* B(:,j)
end
end
for i=1:size(B,2)
TMP=B(:,i);
TMP=TMP./(sqrt(sum(TMP.^2)));
B(:,i)=TMP;
end
end
Find the Normalzied orthogonal basis:
[B] = GramSchmidt_N(A)
Therefore, the Normalized orthogonal Basis (Q) is:
\[\left[ \begin{array} { c c c c } { - 1 / 2 } & { 1 / 2 } & { 1 / \sqrt { 3 } } & { 0 } \\ { 1 / 10 } & { 1 / 2 } & { 0 } & { 1 / \sqrt { 2 } } \\ { - 3 / 10 } & { - 1 / 2 } & { 1 / \sqrt { 3 } } & { 0 } \\ { 4 / 5 } & { 0 } & { 1 / \sqrt { 3 } } & { 0 } \\ { 1 / 10 } & { 1 / 2 } & { 0 } & { - 1 / \sqrt { 2 } } \end{array} \right]\]
Step-2:
Find the matrix R using following formula:\[Q ^ { T } A = R\]
R=Q'*A
Thus, matrix R is: \[\left[ \begin{array} { c c c c } { 20 } & { - 20 } & { - 10 } & { 10 } \\ { 0 } & { 6 } & { - 8 } & { - 6 } \\ { 0 } & { 0 } & { 6 \sqrt { 3 } } & { - 3 \sqrt { 3 } } \\ { 0 } & { 0 } & { 0 } & { 5 \sqrt { 2 } } \end{array} \right]\]
We are given with a matrix: \[A = \left[ \begin{array} { r r r r } { - 10 } & { 13 } & { 7 } & { - 11 } \\ { 2 } & { 1 } & { - 5 } & { 3 } \\ { - 6 } & { 3 } & { 13 } & { - 3 } \\ { 16 } & { - 16 } & { - 2 } & { 5 } \\ { 2 } & { 1 } & { - 5 } & { - 7 } \end{array} \right]\]We have to find QR factorization for the above matrix:
Step-1:
Enter the matrix A in MATLAB.
>> A=[-10 13 7 -11; 2 1 5 3; -6 3 13 -3; 16 -16 -2 5; 2 1 -5 -7];
The required function:
function [B] = GramSchmidt_N(A)
[m,n] = size(A);
[U, jb] = rref(A);
x = length(jb);
B = zeros(m,x);
for i = 1:x
C(:,i)= A(:,(jb(i)));
end
B=C;
for i = 2:x
for j = 1:i-1
B(:,i) = B(:,i) - dot(C(:,i),B(:,j))/dot(B(:,j),B(:,j))* B(:,j)
end
end
for i=1:size(B,2)
TMP=B(:,i);
TMP=TMP./(sqrt(sum(TMP.^2)));
B(:,i)=TMP;
end
end
Find the Normalzied orthogonal basis:
[B] = GramSchmidt_N(A)
Therefore, the Normalized orthogonal Basis (Q) is:
\[\left[ \begin{array} { c c c c } { - 1 / 2 } & { 1 / 2 } & { 1 / \sqrt { 3 } } & { 0 } \\ { 1 / 10 } & { 1 / 2 } & { 0 } & { 1 / \sqrt { 2 } } \\ { - 3 / 10 } & { - 1 / 2 } & { 1 / \sqrt { 3 } } & { 0 } \\ { 4 / 5 } & { 0 } & { 1 / \sqrt { 3 } } & { 0 } \\ { 1 / 10 } & { 1 / 2 } & { 0 } & { - 1 / \sqrt { 2 } } \end{array} \right]\]
Step-2:
Find the matrix R using following formula:\[Q ^ { T } A = R\]
R=Q'*A
Thus, matrix R is: \[\left[ \begin{array} { c c c c } { 20 } & { - 20 } & { - 10 } & { 10 } \\ { 0 } & { 6 } & { - 8 } & { - 6 } \\ { 0 } & { 0 } & { 6 \sqrt { 3 } } & { - 3 \sqrt { 3 } } \\ { 0 } & { 0 } & { 0 } & { 5 \sqrt { 2 } } \end{array} \right]\]