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With the definition:

If $A$ is any subset of $\mathbb{R}^n$, its orthogonal complement, written $A^\perp$, is the set of all vectors $y \in \mathbb{R}^n$ that are perpendicular to every vector in $A$

I have to show these rules:

(a) If $A \subset B$, then $A^\perp \supset B^\perp$.

(b) $A^\perp$ is a subspace (whether A is or not)

(c) $[A]^\perp = A^\perp$ ([A] means span of A)

I know to show something is a subspace of something I just need so show two points that are contained are associative and distributive (x and y in the set then x + y in set and for some scalar c xc is in the set), and similarly the span contains every vector of the form $a_n*v_n$ where a is a real number and v is a vector. I just am having trouble conceptualizing what an orthogonal complement is and how to show these properties.

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  • $\begingroup$ You statements after "I know" are pretty jumbled. I suggest that you review other proofs that certain things are subsets of others, so that you can clearly express the concept before trying to use it in a proof yourself. $\endgroup$ – John Hughes Oct 8 '14 at 18:56
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It may help to start these problems by working through them under the case $A = \{v\}$ for some vector $v$.

For a), show that if $x \in B^\perp$, then $x \in A^\perp$.

You have the right idea for b).

c) is similar to b).

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Hints:

$$\begin{align*}a)&\;\;x\in B^\perp\implies b\cdot x=0\;,\;\forall b\in B\,.\;\text{In particular}\;\;a\cdot x=0\;\;\forall\,a\in A\subset B\implies x\in A^\perp\\{}\\b)&\;\;0\in A^\perp\;\;\text{and}\;\;x,y\in A^\perp\implies 0=x\cdot a-y\cdot a=(x-y)\cdot a\;,\;\;\forall\,a\in A\end{align*}$$

I leave the last one for you, which should now be easier.

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