Solution D.C21 Chris Black

The subspace $W$ can be written as \begin{align*} W &= \setparts{\colvector{a + b\\ a + c\\a + d \\ d}}{a, b, c, d \in\complexes}\\ &= \setparts{ a\colvector{1\\1\\1\\0} + b\colvector{1\\0\\0\\0} + c\colvector{0\\1\\0\\0} + d\colvector{0\\0\\1\\1} } {a, b, c, d \in\complexes}\\ &= \spn{\set{ \colvector{1\\1\\1\\0}, \colvector{1\\0\\0\\0}, \colvector{0\\1\\0\\0}, \colvector{0\\0\\1\\1} }} \end{align*} Since the set of vectors $\set{ \colvector{1\\1\\1\\0},\, \colvector{1\\0\\0\\0},\, \colvector{0\\1\\0\\0},\, \colvector{0\\0\\1\\1} }$ is a linearly independent set (why?), it forms a basis of $W$. Thus, $W$ is a subspace of $\complex{4}$ with dimension 4 (and must therefore equal $\complex{4}$).