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Expert-verified Found in: Page 331 ### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384 # Use the Gram–Schmidt process as in Example 2 to produce an orthogonal basis for the column space ofA = \left( {\begin{aligned}{{}{r}}{ - 10}&{13}&7&{ - 11}\\2&1&{ - 5}&3\\{ - 6}&3&{13}&{ - 3}\\{16}&{ - 16}&{ - 2}&5\\2&1&{ - 5}&{ - 7}\end{aligned}} \right)

The orthogonal basis is,W = \left\{ {\left( {\begin{aligned}{{}{r}}{ - 10}\\2\\{ - 6}\\{16}\\2\end{aligned}} \right),\left( {\begin{aligned}{{}{r}}3\\3\\{ - 3}\\0\\3\end{aligned}} \right),\left( {\begin{aligned}{{}{r}}6\\0\\6\\6\\0\end{aligned}} \right),\left( {\begin{aligned}{{}{r}}0\\5\\0\\0\\{ - 5}\end{aligned}} \right)} \right\}.

See the step by step solution

## Step 1: $$QR$$ factorization of a Matrix

A matrix with order $$m \times n$$ can be written as the multiplication of an upper triangular matrix $$R$$ and a matrix $$Q$$ which is formed by applying the Gram–Schmidt orthogonalization process to the $${\rm{col}}\left( A \right)$$.

The matrix $$R$$ can be found by the formula $${Q^T}A = R$$.

By applying Gram-Schmidt orthogonal process, we can determine the orthogonal basis for the column space of $$A$$

## Step 2: Finding the matrix $$R$$

Given that, A = \left( {\begin{aligned}{{}{r}}{ - 10}&{13}&7&{ - 11}\\2&1&{ - 5}&3\\{ - 6}&3&{13}&{ - 3}\\{16}&{ - 16}&{ - 2}&5\\2&1&{ - 5}&{ - 7}\end{aligned}} \right).

Now with the help of MATLAB, we shall find the orthogonal basis of the column space

MATLAB Command:

Enter 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(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) = C(:,i)- dot(C(:,i),B(:,j))/dot(B(:,j),B(:,j))* B(:,j) ;

end

end

end

Find the orthogonal basis:

(B) = GramSchmidt(A)

\begin{aligned}{}B &= \\\begin{aligned}{{}{r}}{ - 10.0000}&{3.0000}&{8.5000}&{ - 7.9620}\\{2.0000}&{3.0000}&{6.5000}&{5.3232}\\{ - 6.0000}&{ - 3.0000}&{11.5000}&{1.1103}\\{16.0000}&0&{ - 2.0000}&{4.2852}\\{2.0000}&{3.0000}&{ - 3.5000}&{ - 8.2510}\end{aligned}\end{aligned}

So, the orthogonal basis isW = \left\{ {\left( {\begin{aligned}{{}{r}}{ - 10}\\2\\{ - 6}\\{16}\\2\end{aligned}} \right),\left( {\begin{aligned}{{}{r}}3\\3\\{ - 3}\\0\\3\end{aligned}} \right),\left( {\begin{aligned}{{}{r}}6\\0\\6\\6\\0\end{aligned}} \right),\left( {\begin{aligned}{{}{r}}0\\5\\0\\0\\{ - 5}\end{aligned}} \right)} \right\}. ### Want to see more solutions like these? 