Question

In exercises 7-10, show that {u₁, u₂} or {u₁,u₂,u₃} is an orthogonal basis for \({\mathbb{R}^2}\) or \({\mathbb{R}^3}\), respectively. Then express x as a linear combination of the u.

7. \[{u_1} = \left[ {\begin{align}2\\{ - 3}\end{align}} \right]\], \[{u_2} = \left[ {\begin{align}6\\4\end{align}} \right]\], and \[x = \left[ {\begin{align}9\\{ - 7}\end{align}} \right]\]

Short answer

The required linear combination is, \[x = 3{u_1} + \frac{1}{2}{u_2}\].

Step-by-step solution

Linear combination definition

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 \(W\).

Check for orthogonality of given vectors

Find \({u_1} \cdot {u_2}\) as follows:

\(\begin{array}{c}{u_1} \cdot {u_2} = \left( 2 \right)\left( 6 \right) + \left( { - 3} \right)\left( 4 \right)\\ = 12 - 12\\ = 0\end{array}\)

Hence, the vectors are orthogonal to each other, as the vectors are non-zero and linearly independent. Therefore, the given set form a basis for \({\mathbb{R}^2}\).

Express x as a linear combination

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

\[\begin{align}{c}x = \left( {\frac{{x \cdot {u_1}}}{{{u_1} \cdot {u_1}}}} \right){u_1} + \left( {\frac{{x \cdot {u_2}}}{{{u_2} \cdot {u_2}}}} \right){u_2}\\ = \left( {\frac{{\left( 9 \right)\left( 2 \right) + \left( { - 7} \right)\left( { - 3} \right)}}{{\left( 2 \right)\left( 2 \right) + \left( { - 3} \right)\left( { - 3} \right)}}} \right){u_1} + \left( {\frac{{\left( 9 \right)\left( 6 \right) + \left( { - 7} \right)\left( 4 \right)}}{{\left( 6 \right)\left( 6 \right) + \left( 4 \right)\left( 4 \right)}}} \right){u_2}\\ = \left( {\frac{{18 + 21}}{{4 + 9}}} \right){u_1} + \left( {\frac{{54 - 28}}{{36 + 16}}} \right){u_2}\\ = \left( {\frac{{18 + 2}}{{4 + 9}}} \right){u_1} + \left( {\frac{{54 - 28}}{{36 + 16}}} \right){u_2}\,\\ = 3{u_1} + \frac{1}{2}{u_2}\end{align}\]

Hence, \[x = 3{u_1} + \frac{1}{2}{u_2}\].