I've got this problem:
Let $H = \left \{ x \in \mathbb{R}^{4} \, \left| \, x_2 - x_3 + 2x_4 = 0 \right. \right \}$
Find, if possible, $a \in \mathbb{R}$ and $S, T$ vector subspaces so that $\dim(S) = \dim(T)$, $S + T^\perp = H$, $S \cap T = \left \langle (1, a, 0, -1) \right \rangle$
What I have is:
- Using the dimension theorem for vector spaces: $\dim(S+T^\perp) = \dim(S) + \dim(T^\perp) - \dim(S \cap T^\perp) = \dim(H)$. Since $H$ is a $\mathbb{R}^{4}$ vector subspace with one equation, $\dim(H) = 3$. So $\dim(S) = 2$, $\dim(T^\perp) = 2$ and $\dim(S \cap T^\perp)=1$.
- If $\dim(T^\perp) = 2$, then $\dim(T)$ must be 2 as well. So I've got $S=\left \langle s_1, s_2 \right \rangle$ and $T=\left \langle t_1, t_2 \right \rangle$
- Let $s_1, s_2$ two linearly independent vectors from subspace $H$. Suppose $s_1 = (0,1,1,0), s_2 = (0,0,2,1)$. Then $S=\left \langle (0,1,1,0),(0,0,2,1) \right \rangle$.
- Let $t_1, t_2$ two linearly independent vectors from subspace $H$. Suppose $t_1 = (0,-2,0,1), t_2=(1,-1,1,1)$. Then $T^\perp=\left \langle (0,-2,0,1),(1,-1,1,1) \right \rangle$
- Because $(T^\perp)^\perp = T \rightarrow T=\left \{ x \in \mathbb{R}^{4} / -2x_2 + x_4 = x_1 - x_2 + x_3 + x_4 = 0 \right \}$
S and T satisfies all the conditions the problem asks. I know how to find $S \cap T$, but I'm a bit disappointed finding $a$. Any suggestion would be appreciated!
Thanks in advance!