Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 3 / 7C / Solution

Call the four given column vectors $v_1,v_2,v_3,v_4$. Direct componentwise calculation givesThe first two vectors are independent: their first two components give the determinant $1\cdot 2-4\cdot 2=-6\ne 0$. 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 matrixIts 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 ${\mathbb{R}}^5/\mathrm{span}\{(1,1,1,1,1{)}^T\}$. 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 $\boxed{8}$. Requiring the image to equal $W$, instead of being contained in $W$, would not define a vector space; the printed containment is essential.