I'm reading "A Modern Introduction to Linear Algebra" by Henry Ricardo. The author starts the book by Vectors (i.e. no notion about matrices yet). The author explained the spanning set very well; however, in the exercises section, the following problem appears
Let $\boldsymbol{u} = \begin{bmatrix} 2\\1\\1\end{bmatrix}$ and
$\boldsymbol{v} = \begin{bmatrix} 2\\t\\2t\end{bmatrix}$. Find all
values of $t$ (if any) for which $\boldsymbol{u}$ and
$\boldsymbol{v}$ span $\mathbb{R}^3$.
The author implicitly solves a system of linear equation in a particular example which is odd since the notion of a matrix has not been introduced yet. But he explicitly emphasizes on the following form to check for a spanning vector
$$
\begin{align}
k_1 \begin{bmatrix} 2\\1\\1\end{bmatrix} +k_2\begin{bmatrix} 2\\t\\2t\end{bmatrix} =
\begin{bmatrix} x\\y\\z\end{bmatrix}
\end{align}
$$
and the objective is to find a relationship between $k_1,k_2$ and $x,y,z$. Obviously two vectors are not enough to expand the entire $\mathbb{R}^3$ but they may span a plane. My solution is
$$
A= \begin{bmatrix} 2&2&x\\1&t&y\\1&2t&z\end{bmatrix}, \text{rref}(A) = \begin{bmatrix} 1&0&
\frac{tx-2y}{2(t-1)} \\0&1&\frac{2y-x}{2(t-1)}\\0&0&2(t-1)z-2ty+tx\end{bmatrix}
$$
where $k_1 = \dfrac{tx-2y}{2(t-1)}$ and $k_2=\dfrac{2y-x}{2(t-1)}$ and
$$
2(t-1)z-2ty+tx = 0.
$$
The values of $t$ are the ones satisfies the above equation. Is this correct? Why the author implicitly suggests a possibility for two vectors to span $\mathbb{R}^3$? Any suggestions?
Best Answer
You rightly said that two vectors can't span a linear space of dimension equal to $3$.
I think that the author is proposing this exercise to prove that indeed, whatever the value of $t$ is, $\{u,v\}$ can't span $\mathbb R^3$ before he introduces the notion of basis.
To do this, an easy way is to notice that the only way for the first coordinate of $k_1 u + k_2 v$ to vanish, is to have $k_1=-k_2=k \in \mathbb R$. In that case
$$k u - k v = \begin{pmatrix} 0\\ k(1-t)\\ k(1-2t)\end{pmatrix}.$$
Now if $t=1$, you won't be able to span a vector having zero as the first coordinate and one as the second. Hence $t \neq 1$ if we want $\{u,v\}$ to span $\mathbb R^3$. With a similar argument, you can prove that $t \neq 1/2$.
But if $1 \neq t \neq 1/2$, then the second coordinate vanishes if and only if the third one vanishes. A contradiction again if we want $\{u,v\}$ to span $\mathbb R^3$.
We are done.