# Proving $(f^{-1}(U))^0 = f^*(U^0)$

Let $V, W$ be two finite-dimensional vector spaces, $f: V\rightarrow W$ a linear map, and $U \subseteq W$ a vector subspace. I'm trying to show that $(f^{-1}(U))^0 = f^*(U^0)$, i.e. that the annihilator of the inverse image of $U$ is the image of the annihilator under the the dual $f^*$ of $f$. $(f^{-1}(U))^0 \supseteq f^*(U^0)$ is easy to prove, but I'm having troubles with the other direction…

(the annihilator for $Y \subseteq X$ vector spaces is defined here as $Y^0 := \{x^* \in X^*\ |\ \forall y \in Y: x^*(y) =0 \}$, with $X^*$ the dual space of $X$; for inner product spaces, $X^0 \cong X^\perp$).

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Is "annihilator" the orthogonal complement (elements $\mathbf{v}$ such that $\langle\mathbf{v},\mathbf{v'}\rangle=0$ for all $\mathbf{v}'\in f^{-1}(U)$? – Arturo Magidin Mar 15 '12 at 22:08
@klickverbot What is the annihilator? – user38268 Mar 15 '12 at 22:09
@ArturoMagidin, Benjamin Lim: Clarified the question. – klickverbot Mar 15 '12 at 22:17

Given a linear functional $v^*\in(f^{-1}(U))^0$ on $V$ that annihilates all of $f^{-1}(U)$, in particular it annihilates the kernel of $f$. Thus we can define a linear functional $w^*$ on $W$ by $w^*(w)=v^*(v)$ using any preimage $v$ of $w$, since $v^*$ has the same value an all preimages of $w$. Then $w^*\in U^0$ and $v^*=f^*(w^*)$, thus $v^*\in f^*(U^0)$, and thus, since $v^*$ was arbitrary, $(f^{-1}(U))^0\subseteq f^*(U^0)$.

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You can define $w^*$ on $f(V)$ using preimages; what if $f(V)\neq W$? – Arturo Magidin Mar 15 '12 at 22:47
@Arturo: Ah, good point :-). In that case, we can find a basis of $f(V)$, extend it to a basis of $\operatorname{span}(f(V),U)$ using vectors from $U$, then extend it to a basis of all of $W$, and define $w^*$ by linear extension from this basis with the basis vectors of $f(V)$ mapped as above, the additional basis vectors from $U$ mapped to $0$ and the remaining basis vectors mapped arbitrarily. – joriki Mar 15 '12 at 23:01

I claim that $(f^*(U^0))^0\subseteq f^{-1}(U)$.

Let $\mathbf{v}\in (f^*(U^0))^0$. Then for every $\mathbf{x}\in f^*(U^0)$, we have $\langle \mathbf{v},\mathbf{x}\rangle = 0$. Therefore, for every $\mathbf{w}\in U^0$, $$\langle f(\mathbf{v}),\mathbf{w}\rangle = \langle \mathbf{v},f^*(\mathbf{w})\rangle = 0,$$ (since $f^{*}(\mathbf{w})\in f^*(U^0)$). As this holds for all $\mathbf{w}\in U^0$, it follows that $f(\mathbf{v})\in (U^0)^0 = U$. Thus, $\mathbf{v}\in f^{-1}(U)$. This proves the inclusion.

Therefore, $$(f^*(U^0))^0 \subseteq f^{-1}(U),$$ hence $$f^*(U^0) = ((f^*(U^0))^0)^0 \supseteq (f^{-1}(U))^0,$$ which shows that $(f^{-1}(U))^0\subseteq f^*(U^0)$, as required.

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Note that I am using finite dimensionality in order to get $f^*(U^0) = ((f^*(U^0))^0)^0$; I think that the equality may be generally false in the infinite dimensional case, which might explain the difficulty in proving the inclusion directly... – Arturo Magidin Mar 15 '12 at 22:39
Unless my proof is flawed, I think it works for the infinite-dimensional case, too. – joriki Mar 15 '12 at 22:44
I think ${(U^0)}^0 \cong U$ only holds in the finite dimensional case. Also – sorry for that – I edited the question removing the requirement that V and W be Euclidean, since I really want to prove the general case (it just happens that the only application I have for it right now deals with Euclidean spaces). I think, though, that the proof might still work with minor modifications, let me check… – klickverbot Mar 15 '12 at 22:44
@klickverbot: Well, $(U^{\perp})^{\perp}=U$ is guaranteed to hold for all subspaces only in the finite dimensional case, so I expect the same holds for annihilators. – Arturo Magidin Mar 15 '12 at 22:53