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I'm reading Lang's: Linear Algebra. There is an exercise:

Let $A_1, A_2$ be vectors in $\Bbb{R}^n$. Show that the set of all vectors $B\in\Bbb{R}^n$ such that $B$ is perpendicular to both $A_1,A_2$ is a subspace.

I did the following: Using the inner product given before:

  • $$\langle B,A_n \rangle\stackrel{!}{=}0 $$

  • $$\langle \alpha B,A_n \rangle\stackrel{?}{=}0 \\ \alpha \langle B,A_n \rangle\stackrel{?}{=}0 \\ \alpha 0\stackrel{!}{=}0$$

  • $$\langle \alpha B+\beta B,A_n \rangle\stackrel{?}{=}0 \\ \ \langle (\alpha +\beta )B,A_n \rangle\stackrel{?}{=}0 \\ (\alpha +\beta ) \langle B,A_n \rangle\stackrel{?}{=}0\\ (\alpha +\beta ) 0\stackrel{!}{=}0$$

I'm not sure if that is correct though. As we're talking about $\Bbb{R}^n$ there could be more than one vector perpendicular to those two, but the only one I have guarantee that is perpendicular is $B$, given by the exercise.

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    I think you have a misunderstanding in place. You showed that the set is closed under multiplication by scalar, but the second one seems confused. You should show that given $B_1$ and $B_2$ in the set, then their sum is also in the set (which is a straightforward computation). Just switch $\alpha B$ to $B_1$ and $\beta B$ to $B_2$. – 2017-02-06
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    Let $V_1$ -set of vectors perepedincular to $A_1$ and $V_2$ -set of vectors perepedincular to $A_2$. We know that $V_1$ and $V_2$ are subspaces, hence $V_1 \cap V_2$ is also subspace and this is the desired set. – 2017-02-06
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    You need to check $\langle\alpha B_1+\beta B_2,A_n\rangle$=0. – 2017-02-06

1 Answers 1

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Alternative way:

You want to show that $B=($ span$(A_1\cup A_2))^\perp =(A_1+ A_2)^\perp=(A_1)^\perp+ (A_2)^\perp$

where the last equality is shown here: orthogonal complement of a sum

Then you would have:

$(A_1)^\perp+ (A_2)^\perp=$span$((A_1)^\perp\cup(A_2)^\perp)$ but a span is a vector subspace, therefore, $B$ is a vector subspace.