# How to demonstrate if H is a subspace?

I have an exercise which is:

Let $u = (1,2,3)$. If $H=\{u\in\mathbb{R}^3 \mid u\cdot v\}=0$, show that $H$ is a subspace of $\mathbb{R}^3$.

I know if $u\cdot v=0$, they are linearly independent so they should generate a subspace and it would be $\mathbb{R}^2$ but I don't know how to demonstrate it, I never did something like this.

Thank you for your help :)

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I think the sentence "If $H$..." is mistyped? –  JohnD Dec 14 '12 at 18:11

I like inner product notation, so I will write $(u|v)$ to mean what you mean by $u.v$. :)

(1) Contains the $0$ vector: Clearly $(u|0)=0$, so $0 \in H$.

(2) Closure under vector addition: Suppose $x,y\in H$. Then $(u|x)=(u|y)=0$. Then by bilinearity of the scalar product for vector spaces over $\mathbb{R}$, we have $(u|x+y)=(u|x)+(u|y)=0+0=0$. Thus $x+y \in H$.

(3) Closure under scalar multiplication: Suppose $x \in H$. Then $(u|x)=0$. By bilinearity again, for any scalar $c$, we have $(u|cx)=c(u|x)=c\cdot 0=0$. Thus $cx\in H$.

Notice that I didn't actually use that $u=(1,2,3)$. This is a special case of the theorem that orthogonal complements are subspaces.

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Suppose that $v_1,v_2\in H$, $\alpha\in\mathbb R$. Then $$u\cdot(v_1+\alpha\,v_2)=u\cdot v_1+\alpha\,u\cdot v_2=0+\alpha\,0=0,$$ so $v_1+\alpha\,v_2\in H$. As $v_1,v_2$, and $\alpha$ were arbitrary, we conclude that $H$ is a subspace.

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Are you sure that problem is exact? A single set can't be sub space unless it is a zero set, or you define another sum and multiplication so $(1,2,3)$ had a zero role at that relation. Excuse me if my answer is not complete enough!

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This is factually incorrect - every subspace is a single set (equipped with additional structure), for that matter every vector space is a single set. –  kigen Dec 14 '12 at 18:55
Yes, I think I can do that but I only have one vector, whats is the other vector $v$ ? –  Andres Dec 14 '12 at 18:15