Question: In exercises 7-10, show that \(\left\{ {{u_1},{u_2}} \right\}\) or \(\left\{ {{u_1},{u_2},{u_3}} \right\}\) is an orthogonal basis for \({\mathbb{R}^2}\) or \({\mathbb{R}^3}\), respectively. Then express \(x\) as a linear combination of the \(u\)’s.

10. \({u_1} = \left( {\begin{array}{*{20}{c}}3\\{ - 3}\\0\end{array}} \right)\), \({u_2} = \left( {\begin{array}{*{20}{c}}2\\2\\{ - 1}\end{array}} \right)\), \({u_3} = \left( {\begin{array}{*{20}{c}}1\\1\\4\end{array}} \right)\) and \(x = \left( {\begin{array}{*{20}{c}}5\\{ - 3}\\1\end{array}} \right)\)

Let the set of vectors \({u_1},.....,{u_p}\) be an orthogonal basis for a subspace \(W\) of \({\mathbb{R}^n}\) and the linear combination is given by \(y = {c_1}{u_1} + ..... + {c_p}{u_p}\) , then the weights in the linear combination are given as \({c_j} = \frac{{y \cdot {u_j}}}{{{u_j} \cdot {u_j}}}\), for each \(y\) in \(W\).

First, find \({u_1} \cdot {u_2}\):

\(\begin{array}{c}{u_1} \cdot {u_2} = \left( 3 \right)\left( 2 \right) + \left( { - 3} \right)\left( 2 \right) + \left( 0 \right)\left( { - 1} \right)\\ = 6 - 6 - 0\\ = 0\end{array}\)

Now, find \({u_2} \cdot {u_3}\):

\(\begin{array}{c}{u_2} \cdot {u_3} = \left( 2 \right)\left( 1 \right) + \left( 2 \right)\left( 1 \right) + \left( { - 1} \right)\left( 4 \right)\\ = 2 + 2 - 4\\ = 0\end{array}\)

And find \({u_1} \cdot {u_3}\):

\(\begin{array}{c}{u_1} \cdot {u_3} = \left( 3 \right)\left( 1 \right) + \left( { - 3} \right)\left( 1 \right) + \left( 0 \right)\left( 4 \right)\\ = 3 - 3 + 0\\ = 0\end{array}\)

Hence, all vectors are orthogonal to each other as the vectors are non-zero and linearly independent. Therefore, the given set form a basis for \({\mathbb{R}^3}\).

The vector\(x\)can be expressed as a linear combination as follows:

\(\begin{array}{c}x = \left( {\frac{{x \cdot {u_1}}}{{{u_1} \cdot {u_1}}}} \right){u_1} + \left( {\frac{{x \cdot {u_2}}}{{{u_2} \cdot {u_2}}}} \right){u_2} + \left( {\frac{{x \cdot {u_3}}}{{{u_3} \cdot {u_3}}}} \right){u_3}\\ = \left( {\frac{{\left( 5 \right)\left( 3 \right) + \left( { - 3} \right)\left( { - 3} \right) + \left( 1 \right)\left( 0 \right)}}{{\left( 3 \right)\left( 3 \right) + \left( { - 3} \right)\left( { - 3} \right) + \left( 0 \right)\left( 0 \right)}}} \right){u_1} + \left( {\frac{{\left( 5 \right)\left( 2 \right) + \left( { - 3} \right)\left( 2 \right) + \left( 1 \right)\left( { - 1} \right)}}{{\left( 2 \right)\left( 2 \right) + \left( 2 \right)\left( 2 \right) + \left( { - 1} \right)\left( { - 1} \right)}}} \right){u_2} + \left( {\frac{{\left( 5 \right)\left( 1 \right) + \left( { - 3} \right)\left( 1 \right) + \left( 1 \right)\left( 4 \right)}}{{\left( 1 \right)\left( 1 \right) + \left( 1 \right)\left( 1 \right) + \left( 4 \right)\left( 4 \right)}}} \right){u_3}\\ = \left( {\frac{{15 + 9 + 0}}{{9 + 9 + 0}}} \right){u_1} + \left( {\frac{{10 - 6 - 1}}{{4 + 4 + 1}}} \right){u_2} + \left( {\frac{{5 - 3 + 4}}{{1 + 1 + 16}}} \right){u_3}\\ = \left( {\frac{{24}}{{18}}} \right){u_1} + \left( {\frac{3}{9}} \right){u_2} + \left( {\frac{6}{{18}}} \right){u_3}\,\\ = \frac{4}{3}{u_1} + \frac{1}{3}{u_2} + \frac{1}{3}{u_3}\end{array}\)

Hence, \(x = \frac{4}{3}{u_1} + \frac{1}{3}{u_2} + \frac{1}{3}{u_3}\).