Is it possible for two vectors to span $\mathbb{R}^3$? 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?
 A: 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.
A: I did not go over your solution for the system of linear equations. However, I think the author just wants you to have a deep thought about when a set of vectors can span a vector field to get a grasp of dimensions. That is probably why he leaves it open whether there are any values for $t$ for which the vectors span $\mathbb{R}^3$. And to answer your initial question. No, it is not possible to span the $\mathbb{R}^3$ with two vectors only. By understanding the concept of a basis you will know that the $\mathbb{R}^3$ has dimension $3$ and thus need at least three linear independent vectors in order to span $\mathbb{R}^3$.
A: You should read the definition of basis of a vector space. Given a vector space you have a basis consisting of vectors call it $\mathcal{B}$  such that any element $v\in V$ can be expressed as a finite linear combination of elements of $\mathcal{B}$. i.e $\exists v_{i}\in\mathcal{B},c_{i}\in\mathbb{F},n\in\mathbb{N},\,1\leq i\leq n$  such that $v=\sum_{i=1}^{n}c_{i}v_{i}$
And the vectors in $\mathcal{B}$ are linearly independent. That is given an arbitrary finite collection of elements of $\mathcal{B}$ ,say $w_{1},w_{2},...w_{m}\in\mathcal{B}$, we must have the following condition,
If $\sum_{i=1}^{m}d_{i}w_{i}=0\implies d_{i}=0\,,\forall 1\leq i\leq m$.
The cardinality of the set $\mathcal{B}$ is called the Dimension of the Vector space  denoted by $\dim$.
Now in the case of $\mathbb{R}^{3}$ over the field $\mathbb{R}$. You have the vectors $\{(1,0,0),(0,1,0),(0,0,1)\}$ form a basis. (Verify this!).
Now you can prove that if a vector space has $\dim(V)=n$ . Then at most $n$ elements of $V$ are linearly independent. That is more than $n$ elements cannot be linearly independent.
Also you can prove that any spanning set of $V$ must have atleast $n$ vectors. That is a set with less than $n$ vectors cannot span the entire space $V$. This is what you need to use.
So yeah, for no value of $t$ the two vectors can span $\mathbb{R}^{3}$ which has dimension $3$.
While studying vector spaces(specially finite dimesnional ones) for the first time remember these two sentences:-
1. The basis is the maximal linearly independent set.
2. The basis is the minimal spanning set.
All this said, the author will probably prove all of these after a few pages of introduction. He is trying to give you the feel of how only finitely many elements of an infinite set can give you the entire space by just using linear combinations.
