# Find a basis for s prep

Q. S is a subspace of R^3 containing only the zero vector. If S is spanned by (1,1,1) and (1,1,-1) what is a basis for S perp?

This is what I have so far -> a+b+c = 0 and a+b-c = 0. 2a+2b = 0

[-b] [-a] [0]

so a basis is [-1] [-1] [0]

I am confused. I am not sure what I am doing. Any help would be appreciated.

Thanks.

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Do you mean $S$ perp (as in perpendicular)? This space: $S^{\perp} = \{ x \in \Bbb{R}^3 \; \mid \; x \cdot y = 0 \text{ for all } y \in S \}$. – Sammy Black Feb 27 '14 at 7:59

I think you mean $$S = \mbox{span}\{(1,1,1)\, (1,1,-1)\},$$ which is not the subspace containing only the zero vector. As far as finding the perp, what you have is almost there. A vector $(a,b,c) \in S^\perp$ must dot product with those two to zero giving the equations you obtained. And indeed, from those you get $2a + 2b = 0$. This means that $b = -a$, so we can choose $a = 1$, so that $b = -1$. Also, note that $c = 0$ because if you subtract the equations, you get $2c = 0 \rightarrow c = 0$. Therefore, $(1,-1,0)$ is a basis for $S^\perp$.

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I can choose any number of a, right? (as long as it turns out to be linearly independent). Here's the exact question: If S is the subspace of R3 containing only the zero vector, what is S prep? If S is spanned by (1, 1, 1), what is S prep ? If S is spanned by (1, 1, 1) and (1, 1, -1), what is a basis for S prep ? – Rami Feb 27 '14 at 8:12
You can choose any number $a$ so that in general the vector is of the form $(a,-a,0)$. Note that this is three different questions: 1. $S = \{0\}$, 2. $S = \mbox{span}\{(1,1,1)\}$, and 3. $S = \mbox{span}\{(1,1,1),(1,1-1)\}$. – Suugaku Feb 27 '14 at 8:16
So, should the basis be (a,b,c) or (-b,-a,c), where c = 0. I am sort of confused. I am referring to my answer on line 4 of the original post. – Rami Feb 27 '14 at 8:25
The basis should be $\{(a,b,0)\}$, with $b = -a$, so $\{(a,-a,0)\}$ is a basis for the third question. – Suugaku Feb 27 '14 at 8:31

If $S$ contains only the zero vector, then $S$ is not spanned by $(1,1,1)$ and $(1,1,-1)$. Also, what does prep stand for?

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S⊥ is the orthogonal subspace of S (pronounced S prep). Could you please explain a little why S cannot be spanned? – Rami Feb 27 '14 at 8:08
If $S$ is spanned by $(1,1,1)$ and $(1,1,-1)$, then it contains $(1,1,1)$. Since $S$ contains only the zero vector, this is a contradiction. – 5xum Feb 27 '14 at 8:12