Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question

1 Answer 1

up vote 0 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.