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If we have a vector space $V$ and subspace $W$, we have that $$\dim(V/W) = \dim V - \dim W.$$

Similarly for the annihilator $W^{\circ}$ we have that

$$\dim W^{\circ} = \dim V - \dim W.$$

What is the isomorphism between these two spaces? Is there an intuitive way to relate the two ideas?

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The natural isomorphism is not between $W^0$ and $V/W$, but between $W^{\circ}$ and $(V/W)^*$. Consider the linear map

\begin{align*} L : W^{\circ} &\to (V/W)^*\\ \varphi &\mapsto \hat{\varphi} \end{align*}

where $\hat{\varphi}(v + W) := \varphi(v)$. Note, the map $\hat{\varphi}$ is well-defined because if $v' + W = v + W$, then $v' = v + w$ for some $w \in W$ so

$$\hat{\varphi}(v' + W) = \varphi(v') = \varphi(v + w) = \varphi(v) + \varphi(w) = \varphi(v) = \hat{\varphi}(v + W).$$

Note that if $\hat{\varphi} = 0$, then for every $v \in V$, $0 = \hat{\varphi}(v + W) = \varphi(v)$, so $\varphi = 0$. Therefore $L$ is injective.

Let $\psi \in (V/W)^*$. Note that $\psi\circ\pi \in V^*$ where $\pi : V \to V/W$ is the natural projection, and $(\psi\circ\pi)(W) = \psi(\pi(W)) = \psi(0) = 0$, so $\psi\circ\pi \in W^{\circ}$. Furthermore

$$\widehat{\psi\circ\pi}(v + W) = (\psi\circ\pi)(v) = \psi(\pi(v)) = \psi(v + W),$$

so $L(\psi\circ\pi) = \psi$. Therefore, $L$ is surjective.

So the map $L : W^{\circ} \to (V/W)^*$, $\varphi \mapsto \hat{\varphi}$ is an isomorphism.

If $V$ (and hence $W$) is finite-dimensional, then we have

$$\dim W^{\circ} = \dim (V/W)^* = \dim V/W = \dim V - \dim W.$$

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