= Solution
Call the four given column vectors $v_1,v_2,v_3,v_4$. Direct componentwise calculation gives
$$
v_3=2v_1+\frac12v_2,\qquad v_4=v_1+v_2.
$$
The first two vectors are independent: their first two components give the determinant $1\cdot2-4\cdot2=-6\ne0$. They form a <basis> of $W$, so $\boxed{\dim W=2}$.
Choose columns $v_1,v_2,0,0,-v_1-v_2$. This gives the <matrix>
$$
\boxed{M=\begin{pmatrix}1&4&0&0&-5\\2&2&0&0&-4\\2&-2&0&0&0\\-1&6&0&0&-5\\1&-2&0&0&1\end{pmatrix}.}
$$
Its <image> is exactly $W$, since all its columns belong to $W$ and its first two span $W$. Its column sum is zero, so $(1,1,1,1,1)^T$ belongs to its <kernel>.
All <linear maps> with the stated kernel condition and image contained in $W$ factor uniquely through the <quotient vector space> $\mathbb R^5/\operatorname{span}\{(1,1,1,1,1)^T\}$. This quotient has dimension four. A <linear map> from it to the two-dimensional $W$ is specified by eight independent real coordinates, so the requested space has dimension $\boxed8$. Requiring the image to equal $W$, instead of being contained in $W$, would not define a vector space; the printed containment is essential.
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