Im reading Peter Lax book and he says: For any subset $S \subset X'$, we define $S^\perp$ as the subset of those vectors in $X$ that are annihilated by every vector in S. This confuses me a bit, shoudent it be every functional in X'' that vanishes on S? Or is this the same thing by identifying those vectors in X?
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
Yes, it should be the functionals. I'll try to give you a more accurate description: Let V be a vector space over a field F, and $$V^{*}$$ be V's dual space (meaning the space of functions from V to F). Let $$ S \subseteq V^{*}$$ be a subset of the dual space, meaning, it's a set of functionals. Let's define $$S^{\perp}$$ to be the set of vectors v in V, such that for all the functionals f in $$S \subseteq V^{*}$$, f(v)=0. Or in formal language: $$ S^{\perp}=\left \{ v \in V: \forall f \in S, f(v)=0 \right \} $$ |
|||
|
|
There are two choices for definition of annihilators in $V^*$, the dual of a vector space $V$. The first is the one you give, $\{v \in V : \alpha(v) = 0 \space \forall \alpha \in S\}$. The second, using the same definition for annihilators in $V^*$ as we do in $V$, $\{\theta \in V^{**}: \theta(\alpha) = 0 \space \forall \alpha \in S\}$. If the space is finite dimensional, then $V$ is naturally isomorphic to $V^{**}$ by the map $v \mapsto\hat {\hat v}$ where $\hat{\hat v}(\alpha) = \alpha(v)$. Under this isomorphism, the two different annihilators are isomorphic. If $V$ is not finite dimensional then $V$ and $V^{**}$ are not generally isomorphic and so the different definitions give genuinely different sets. |
|||
|
|
