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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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  • $\begingroup$ I think the sentence "If $H$..." is mistyped? $\endgroup$
    – JohnD
    Dec 14, 2012 at 18:11

4 Answers 4

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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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    $\begingroup$ This is factually incorrect - every subspace is a single set (equipped with additional structure), for that matter every vector space is a single set. $\endgroup$ Dec 14, 2012 at 18:55
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By the Subspace Theorem you only need to establish the zero vector is in there as well as closure under addition and scalar multiplication. Can you do that?

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  • $\begingroup$ Yes, I think I can do that but I only have one vector, whats is the other vector $v$ ? $\endgroup$
    – Andres
    Dec 14, 2012 at 18:15
  • $\begingroup$ It's not just nonemptiness. You need to show that the zero vector is in the set (which implies nonempty, but of course nonempty isn't enough to imply inclusion of the zero vector). $\endgroup$ Dec 14, 2012 at 18:57
  • $\begingroup$ @JohnD This is certainly true! But I think the Subspace Theorem states it specifically as "0 is in the set" because it's like the corresponding theorem for proving subgroups: you need to show the identity is in the set. Besides "0 is in the set" is a clear and concrete starting point for proofs, and also comes up a lot when you want to show a set is not a subspace, so I prefer it to "the set is nonempty and closed under scalar multiplication". $\endgroup$ Dec 14, 2012 at 19:23

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