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Let $A, B$ be subspaces. I want to show the following: $A \subseteq B \implies B^{\perp} \subseteq A^{\perp}$

Is the following a legitimate proof?

If $x \in A$, then $x \in B$ since $A \subseteq B$. Assume $x$ is not the zero vector. Then because $x \in A$, it must be the case that $x \notin A^{\perp}$. Similarly because $x \in B$ that means $x \notin B^{\perp}$. So we have shown that $x \notin A^{\perp} \implies x \notin B^{\perp}$.

Now by the contrapositive, we see that $x \in B^{\perp} \implies x \in A^{\perp}$. So $B^{\perp} \subseteq A^{\perp}$

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Your argument is not correct! $x \not\in A$ do not imply that $x \in A^\perp$. – kjetil b halvorsen Oct 10 '12 at 23:27
@kjetil: I don't see where that's asserted in the question. – joriki Oct 10 '12 at 23:31
It seems tyo be there implicitely. – kjetil b halvorsen Oct 11 '12 at 17:09
up vote 2 down vote accepted

It's not generally true that $x\in A$ implies $x\notin A^\perp$, for instance in $\mathbb F_2^n$.

The proof is simpler than that and doesn't rely on any specifics of orthogonal complements. Since the orthogonal complement of $B$ is the set of all vectors that are orthogonal to all vectors in $B$, if we remove some elements from $B$ we've just made the requirement less stringent, so we can't have excluded any vectors from the orthogonal complement that were in it before; thus the orthogonal complement of a subset must be a superset of the orthogonal complement.

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Thank you for the nice intuitive argument. Also what is $\mathbb{F}^n_2$ – Student Oct 10 '12 at 23:34
@Student: $\mathbb F_2$ is the field with two elements, the smallest finite field, and $\mathbb F_2^n$ is the $n$-dimensional vector space over $\mathbb F_2$ of $n$-tuples with elements in $\mathbb F_2$. I suspect that you were only thinking of vector spaces over $\mathbb R$ or perhaps $\mathbb C$, but since your conclusion holds in all vector spaces equipped with a bilinear form, over arbitrary fields, the proof should ideally work for all fields and not rely on specifics of $\mathbb R$ and $\mathbb C$. – joriki Oct 10 '12 at 23:44
But if my proof was for just $\mathbb{R}$ or $\mathbb{C}$ would it be ok? – Student Oct 10 '12 at 23:45
@Student: No, it wouldn't. You assume that $x\in A$ and then try to derive an implication with premise $x\notin A^\perp$. But that implication has to hold for all $x\notin A^\perp$, not just for $x\in A$, so you can't prove it starting from $x\in A$. (The reason I mentioned $\mathbb R$ is just that $(x\ne0\land x\in A)\Rightarrow x\notin A^\perp$ does hold in $\mathbb R^n$.) – joriki Oct 10 '12 at 23:48
I get it now. Thanks! – Student Oct 10 '12 at 23:52

Absolutely not.

If you want to prove about sets that $X\subseteq Y$, then it's usually going as:

  1. suppose we are given $x\in X$
  2. Then try to prove $x$ satisfies the criterium for $Y$ (i.e. $x\in Y$).

Now: suppose we are given $u\in B^\perp$ (what does it mean? $u$ is orthogonal to the whole $B$), then conclude that $u\in A^\perp$, of course, using the hypothesis $A\subseteq B$.

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