Question

Let \(u = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\) and \(v = \left[ {\begin{array}{*{20}{c}}2\\1\end{array}} \right]\). Show that \(\left[ {\begin{array}{*{20}{c}}h\\k\end{array}} \right]\) is in Span \(\left\{ {u,v} \right\}\) for all \(h\) and\(k\).

Short answer

\(\left[ {\begin{array}{*{20}{c}}h\\k\end{array}} \right]\) is in span\(\left\{ {u,v} \right\}\) for all \(h\) and \(k\).

Step-by-step solution

Determine the objective to be shown

You want to show that \(\left[ {\begin{array}{*{20}{c}}h\\k\end{array}} \right]\) is in span\(\left\{ {u,v} \right\}\) for all \(h\) and \(k\). It is enough to depict that there exists \(x\) and \(y\) in \(\mathbb{R}\) such that\(xu + yv = \left[ {\begin{array}{*{20}{c}}h\\k\end{array}} \right]\).

Convert the linear combination to the system

Substitute \(u = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\) and \(v = \left[ {\begin{array}{*{20}{c}}2\\1\end{array}} \right]\) in \(xu + yv = \left[ {\begin{array}{*{20}{c}}h\\k\end{array}} \right]\) as shown below:

\(\begin{array}{c}x\left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right] + y\left[ {\begin{array}{*{20}{c}}2\\1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}h\\k\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{2x}\\{ - x}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{2y}\\y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}h\\k\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{2x + 2y}\\{ - x + y}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}h\\k\end{array}} \right]\end{array}\)

This implies that the system of equations is:

\(\begin{array}{c}2x + 2y = h\\ - x + y = k\end{array}\)

Determine the consistency of the system

The row echelon form of the augmented matrix of this system is given as follows:

Apply row operation \({R_1} \to \frac{{{R_1}}}{2}\) to \(\left[ {\begin{array}{*{20}{c}}2&2&h\\{ - 1}&1&k\end{array}} \right]\) to get

\(\left[ {\begin{array}{*{20}{c}}1&1&{\frac{h}{2}}\\{ - 1}&1&k\end{array}} \right]\)

Apply \({R_2} \to {R_2} + {R_1}\) in the above matrix to get

\(\left[ {\begin{array}{*{20}{c}}1&1&{\frac{h}{2}}\\0&2&{k + \frac{h}{2}}\end{array}} \right]\)

This implies that the system is consistent.

Equivalent system of the original system

The equivalent system of equations using the echelon form is:

\(\begin{array}{c}x + y = \frac{h}{2}\\2y = k + \frac{h}{2}\end{array}\)

Solve x and y

Solve \(x\) and \(y\) in terms of \(h\) and \(k\) as follows:

\(y = \frac{1}{2}\left( {k + \frac{h}{2}} \right)\)

Substitute \(y = \frac{1}{2}\left( {k + \frac{h}{2}} \right)\) in \(x + y = \frac{h}{2}\) to get

\(\begin{array}{c}x + \frac{1}{2}\left( {k + \frac{h}{2}} \right) = \frac{h}{2}\\x + \frac{k}{2} + \frac{h}{4} = \frac{h}{2}\\x = \frac{h}{2} - \frac{h}{4} - \frac{k}{2}\\x = \frac{h}{4} - \frac{k}{2}\end{array}\)

Conclusion

This implies that there exists \(x\) and \(y\) in \(\mathbb{R}\) such that \(xu + yv = \left[ {\begin{array}{*{20}{c}}h\\k\end{array}} \right]\) for all \(h\) and \(k\).

Hence, \(\left[ {\begin{array}{*{20}{c}}h\\k\end{array}} \right]\) is in span\(\left\{ {u,v} \right\}\) for all \(h\) and \(k\) shown.