Exercise 11

As defined in this section, the span of a set of vectors is generated by taking all possible linear combinations of those vectors. This exercise will demonstrate the fact that the span can also be realized as the solution space to a linear system.

We will consider the vectors $$\vvec_1=\threevec{1}{0}{-2}, \vvec_2=\threevec{2}{1}{0}, \vvec_3=\threevec{1}{1}{2}$$

1. Is every vector in $\real^3$ in $\laspan{\vvec_1,\vvec_2,\vvec_3}$? If not, describe the span.

2. To describe $\laspan{\vvec_1,\vvec_2,\vvec_3}$ as the solution space of a linear system, we will write $$\bvec=\threevec{a}{b}{c}$$. If $\bvec$ is in $\laspan{\vvec_1,\vvec_2,\vvec_3}$, then the linear system corresponding to the augmented matrix $$\left[\begin{array}{rrr|r} 1 & 2 & 1 & a \\ 0 & 1 & 1 & b \\ -2& 0 & 2 & c \\ \end{array}\right]$$ must be consistent. This means that a pivot cannot occur in the rightmost column. Perform row operations to put this augmented matrix into a triangular form. Now identify an equation in $a$, $b$, and $c$ that tells us when there is no pivot in the rightmost column. The solution space to this equation describes $\laspan{\vvec_1,\vvec_2,\vvec_3}$.

3. In this example, the matrix formed by the vectors $\left[\begin{array}{rrr} \vvec_1&\vvec_2&\vvec_3 \\ \end{array}\right]$ has two pivot positions. Suppose we were to consider another example in which this matrix had had only one pivot position. How would this have changed the linear system describing $\laspan{\vvec_1,\vvec_2,\vvec_3}$?