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# Is this vector in the span? Case of no

## Problem

Let

$$
\bu_1 = \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix},\
\bu_2 = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix},\ \text{ and }
\bb = \begin{bmatrix} 5 \\ 5 \\ 6 \\ 6 \end{bmatrix}.
$$

and $S = \{\bu_1, \bu_2\}$. Is $\bb$ in $\vspan(S)$?

## Thought

Recall that $\vspan(S)$ contains all vectors of the form $c_1\bu_1 + c_2\bu_2$ for some $c_1,c_2\in\mathbb{R}$. By definition, $\bb\in\vspan(S)$ if and only if there are $c_1, c_2\in\mathbb{R}$ such that $c_1\bu_1 + c_2\bu_2 = \bb$. Therefore, the problem is equivalent to whether

$$
\begin{aligned}
c_1 + c_2 &= 5, \\
c_1 + 2c_2 &= 7, \\
c_1 + 3c_2 &= 9, \\
c_1 + 4c_2 &= 11 \\
\end{aligned}
$$

has a solution or not.

## Sample answer

The equation $c_1\bu_1 + c_2\bu_2 = \bb$ is equivalent to

$$
\begin{aligned}
c_1 + c_2 &= 5, \\
c_1 + 2c_2 &= 5, \\
c_1 + 3c_2 &= 6, \\
c_1 + 4c_2 &= 6. \\
\end{aligned}
$$

By substracting the first equation from the remaining equations, we have

$$
\begin{aligned}
c_1 + c_2 &= 5, \\
c_2 &= 0, \\
2c_2 &= 1, \\
3c_2 &= 1. \\
\end{aligned}
$$

Thus, there is no solution for this system since it is impossible for $c_2 = 0$ and $2c_2 = 1$. As a result, we know $\bb$ is not in $\vspan(S)$.

## Note

One may visualize $\bb$ and $\vspan(S)$ as the picture below.


*This note can be found at Course website > Learning resources.*