# Finding vector subspaces

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!

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## 1 Answer

Suppose there are such $S$, $T$, and $a$. Since $v := (1, a, 0, -1) \in S \cap T \subseteq S + T^{\perp} = H$, one has $a - 0 + 2\cdot(-1) = 0$, so $a = 2$.

I didn't understand your argument to deduce the dimensions of $S$ and $T$, so I'll give one myself:

Because $\dim (S\cap T) = 1$, you can conlude $1 \leq \dim T = \dim S \leq 2$ (If $\dim T = \dim S > 2$, in 4-dimensional space they met in dimension $>1$, which follows from the dimesion formula you have given). But the equation of $H$ says that $w := (0, 1, -1, 2)$ and $v = (1, 2, 0, -1) \in T$ are orthogonal: So if $\dim T = 1$, then $w \in T^{\perp} \subseteq H$, but $w \notin H$. This cannot be. Therefore $\dim S = \dim T = 2$, as you said.

So, now all you have to do, is complement $v$ with vectors $u_1, u_2$, orthogonally to $v$, to a base of $H = \langle v, u_1, u_2 \rangle$, take that completion as a base of $T^{\perp} = \langle u_1, u_2 \rangle$, and set $S = \langle v, u_1 \rangle$. Now $v \in T \cap S$, but since $u_1 \notin T$, the spaces $S$ and $T$ meet in $\langle v \rangle$.

I think you can take $u_1 = (2, -1, -1, 0)$ and $u_2 = (0, -1, -5, -2)$, they both should be orthogonal to $v$ and $w$.

I also don't understand, how you came up with your $T$ and $S$ though, it seems it might not be working, but I haven't looked into it, to be honest.

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