Suppose W = $ span\{w1,w2,w3,w4\}$ = span $ \left\{ \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ 2 \\ 1 \\ 3 \end{pmatrix}\ \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \\ 1 \end{pmatrix}\right\} $ $ \subseteq $ $R^5 .$
The RREF of $ \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 0 & 1 \\ 1 & 0 & 2 & 0 \\ 1 & 1 & 1 & 0 \\ 1 & -1 & 3 & 1 \end{pmatrix}$ is $ \begin{pmatrix} 1 & 0 & 2 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}.$
Is the dim(W) = 3?
When I did row reduction, I found that w1, w2, w3 except w4 can be expressed as a linear combination of the other vectors $ w_i$ in the set excluding itself, (e.g w2 expressed as a linear combination of w1,3,4), doesn't this mean that the number of vectors in the basis is 1? So, why would the dim(W) = 3? Please correct me if I am wrong, thank you.
There is some confusion here. Consider, for instance, the vectors $(1,0,0)$, $(0,1,0)$, and $(1,1,0)$ from $\Bbb R^3$. Each of them is a linear combination of the other two. However, they span $\{(x,y,0)\mid x,y\in\Bbb R\}$, which is $2$-dimensional, not $1$-dimensional.
Your case is similar: the dimension of $W$ is $3$, not $1$.
A vector space $V$ distinct from $\{0\}$ is $1$-dimensional if and only if, given $v,w\in V\setminus\{0\}$, there is some scalar $\lambda$ such that $v=\lambda w$. That's clearly not what happens with $W$.