Let $\mathbf{u}=\left[\begin{array}{r}15 \\ -1\end{array}\right]$ and $\mathbf{v}=\left[\begin{array}{r}15 \\ 1\end{array}\right]$. Show that $\left[\begin{array}{l}h \\ c\end{array}\right]$ is in $\operatorname{Span}\{\mathbf{u}, \mathbf{v}\}$ for all $h$ and $c$
Solution
Step 1 ：Given vectors \(\mathbf{u} = \left[\begin{array}{r}15 \\ -1\end{array}\right]\) and \(\mathbf{v} = \left[\begin{array}{r}15 \\ 1\end{array}\right]\), we are asked to show that any 2D vector \(\left[\begin{array}{l}h \\ c\end{array}\right]\) is in the span of \(\mathbf{u}\) and \(\mathbf{v}\).
Step 2 ：This is equivalent to saying that any 2D vector can be expressed as a linear combination of the vectors \(\mathbf{u}\) and \(\mathbf{v}\).
Step 3 ：To prove this, we can set up a system of linear equations where the coefficients of \(\mathbf{u}\) and \(\mathbf{v}\) are variables, and the components of the arbitrary vector \(\left[\begin{array}{l}h \\ c\end{array}\right]\) are the constants.
Step 4 ：The system of equations is as follows: \(15a + 15b = h\) and \(-a + b = c\).
Step 5 ：Solving this system of equations, we find that \(a = -\frac{c}{2} + \frac{h}{30}\) and \(b = \frac{c}{2} + \frac{h}{30}\).
Step 6 ：This means that for any values of \(h\) and \(c\), we can find values of \(a\) and \(b\) such that the linear combination of \(\mathbf{u}\) and \(\mathbf{v}\) equals \(\left[\begin{array}{l}h \\ c\end{array}\right]\).
Step 7 ：\(\boxed{\text{Therefore, for all } h \text{ and } c, \left[\begin{array}{l}h \\ c\end{array}\right] \text{ is in } \operatorname{Span}\{\mathbf{u}, \mathbf{v}\}}\)
