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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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    It seems tyo be there implicitely.2012-10-11

3 Answers 3

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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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    I get it now. Thanks!2012-10-10
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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 $A\subseteq B$.