The posed questions are:
1) Can every vector in ${\Bbb R}^4$ be written as a linear combination of the columns of a matrix $A$?
2) Do the columns of $A$ span ${\Bbb R}^3$?
I've reduced the $4\times 4$ matrix $A$ down to RREF. The RREF matrix is shown here.
\begin{bmatrix}1&0&-4&0\\0&1&3&0\\0&0&0&1\\0&0&0&0\end{bmatrix}
The following theorem is what I used to answer the first question:
Let $A$ be an $m \times n$ matrix: the following are all true or all false.
$1$. For each $b$ in ${\Bbb R}^m$, $Ax = b$ has a solution
$2$. Each $b$ in ${\Bbb R}^m$, is a linear combo of the columns in $A$.
$3$. The columns of A span ${\Bbb R}^m$
$4$. $A$ has a pivot position in every row
So to answer the first question above, there are only $3$ pivots so $4$ is false in the theorem and therefore $2$ is false. But the theorem only accounts for ${\Bbb R}^m$, not ${\Bbb R}^{m-1}$. To answer the second question above, can I extrapolate that because ${\Bbb R}^3$ is a subspace of ${\Bbb R}^4$, and there are $3$ pivots, that then the above theorem holds true?
Clearly the columns do not span $\mathbb{R}^3$ because, if it span $\mathbb{R}^3$, then we can find $a_i \in \mathbb{R}$ such that
$$\sum_{i=1}^4 a_iA_i=\begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix} $$
where $A_i$ denotes the $i$-th column of $A$.
The size of LHS is $4 \times 1$ and the size of the matrix of RHS is $3 \times 1$. By definition of equality in matrices, they cannot be equal since the size is different.