Problem

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 $c$

Answer

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Answer

\(\boxed{\text{Therefore, for all } h \text{ and } c, \left[\begin{array}{l}h \\ c\end{array}\right] \text{ is in } \operatorname{Span}\{\mathbf{u}, \mathbf{v}\}}\)

Steps

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 :\(\boxed{\text{Therefore, for all } h \text{ and } c, \left[\begin{array}{l}h \\ c\end{array}\right] \text{ is in } \operatorname{Span}\{\mathbf{u}, \mathbf{v}\}}\)

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