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?


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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.