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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?

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Berci makes a very good suggestion... Otherwise, you need to add $(-2,1,0,1)$ to $H$, but then $H$ is probably just $R^4$.... –  N. S. Oct 10 '12 at 1:49

2 Answers 2

up vote 1 down vote accepted

Since it's already been noticed there must be a typo somewhere, I started looking at plausible places where the mistake could have occurred.

The thing that jumps out is that if the sign on $x_4$ in the equation for $\mathbb{S}$ is supposed to be positive, then $\mathbb{S}\cap\mathbb{T}\subseteq \mathbb{H}$.

Making that change, you find $\mathcal{B}=\{(1,0,-1,-1),(0,1,0,-1)\}$ is a basis for $\mathbb{S}\cap\mathbb{T}$. Then $(0,1,-1,1)$ is in $\mathbb{S}^\perp +\mathbb{T}^\perp$ and is linearly independent with the members of $\mathcal{B}$, and since $\mathbb{H}$ is three dimensional, we have that $\mathbb{W}=\mathbb{S}^\perp$ would be a solution in that case.

Similarly you could suppose that $2x_4$ should actually be 0 in the equation for $\mathbb{H}$, but that seems like a less likely mistake. Really there are probably lots of solutions for perturbations of this problem.

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You are totally right. Since $\Bbb S\cap\Bbb T\nsubseteq \Bbb H$, such a $\Bbb W$ cannot exists.

Perhaps it's just a typo, maybe it's $\Bbb S\cap\Bbb T\cap\Bbb H$, or the RHS is the whole $\Bbb R^n$...

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$\Bbb W\oplus (\Bbb S\cap \Bbb T \cap \Bbb H)=\Bbb H$ should be the natural choice here. Wouldn't that make the problem always have a solution, no matter what $S,T,H$ are? –  N. S. Oct 10 '12 at 1:47
    
What I'm thinking is the typo might be in the eqn of $\Bbb H$. –  Pedro Tamaroff Oct 10 '12 at 1:55

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