Consider the following subspaces of $\mathbb R^4$.
$\eqalign{ & \Bbb S = \left\{ {x\in\mathbb R^4:{x_2} - {x_3} - {x_4} = 0} \right\} \cr & \Bbb T = \left\{ {x\in\mathbb R^4:{x_1} + {x_2} + {x_4} = 0} \right\} \cr &\Bbb H = \left\{ {x\in\mathbb R^4:{x_1} + 2{x_2} - {x_3} + 2{x_4} = 0} \right\} \cr}$
Find $\mathbb W$, a subspace, such that
$\mathbb W\subseteq \Bbb S^\perp+\Bbb T^\perp$ $\Bbb W\oplus (\Bbb S\cap \Bbb T)=\Bbb H$
My work: From the equations defining the subspaces, one gets the following:
Let $\eqalign{ & \Bbb S = \left\langle {\left( {1,0,0,0} \right);\left( {0,1,1,0} \right);\left( {0,1,0,1} \right)} \right\rangle \cr & \Bbb T = \left\langle {\left( {0,0,1,0} \right);\left( {1,0,0, - 1} \right);\left( {0,1,0, - 1} \right)} \right\rangle \cr & \Bbb H = \left\langle {\left( {0, - 1,0,1} \right);\left( {1,0,1,0} \right);\left( {0,1,2,0} \right)} \right\rangle \cr} $
Then, one gets:
$\eqalign{ & \Bbb S^\perp = \left\langle {\left( {0,1, - 1, - 1} \right)} \right\rangle \cr & \Bbb T^\perp = \left\langle {\left( {1,1,0,1} \right)} \right\rangle \cr} $
so that
${{\Bbb S}^ \bot } + {{\Bbb T}^ \bot } = \left\langle {\left( {0,1, - 1, - 1} \right);\left( {1,1,0,1} \right)} \right\rangle {\text{ }}$
Finally, for $ \Bbb S\cap \Bbb T$, we consider the system
$\begin{cases}{x_1} + {x_2} + {x_4}=0\\{x_2} - {x_3} - {x_4} = 0\end{cases}$ which gives
$\Bbb S \cap \Bbb T = \left\langle {\left( { - 1,1,1,0} \right)\left( { - 2,1,0,1} \right)} \right\rangle $
My problem is that $\left( { - 2,1,0,1} \right) \notin \Bbb H$; so how can it be that there exists a $\Bbb W$ such that $\Bbb W\oplus (\Bbb S\cap \Bbb T)=\Bbb H$? Can you suggest any modification to $\Bbb H$ to make the problem work out?