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If $W$ is a subspace of $V$, then a well known fact is $$(V/W)' \cong W^0.$$

I'm trying to understand this intuitively/geometrically, perhaps working in a simpler space of $V = \mathbb{R}^2$ and $W$ the $x$-axis.

Intuitively, I understand $V / W$ as the set of all lines parallel to the $x$-xis, and, based on Why do we care about dual spaces?, the dual space is all the various ways you could "summarize" those lines with a single scalar.

On the other hand, I intuitively understand $W^0$ as the set of all ways you can "collapse" the $x$-axis into $0$. But I am struggling to articulate the intuition behind why the two are related.

greg115
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    Suppose $L \in W^0$. It feels tempting to associate $L$ with the function $T(v + W) = L(v)$, which is an element of $(V/W)’$. The functional $L$ views elements of $W$ as negligible, it doesn’t “see” them at all, so it seems natural to think of $L$ as a function on $V/W$, which is the space that results from viewing elements of $W$ as negligible. – littleO Oct 12 '22 at 05:11
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    @littleO Amazing! I got it now – greg115 Oct 12 '22 at 05:17

2 Answers2

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From your intuition in $\mathbb{R}^2$, if $W$ is the $x$ axis, then $V/W$ is isomorphic to the $y$ axis, since every line parallel to the $x$ axis intersects the $y$ axis at a single point (it's height). Now, the dual of $V/W$ is isomorphic to the dual of the $y$ axis, and the dual of the $y$ axis is just by definition all the ways to orthogonaly project into the $y$ axis (because $\mathbb{R}^n$ has an inner product). Clearly, orthogonal projection into the $y$ axis annihilates any $x$ component, and visually it should be clear that those are the only such maps.

*This is all up to isomorphism, and specifically in $\mathbb{R}^2$, but the intuition should carry over easily.

Carl Chaanin
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I'm late to the party but wanted to add that personally I find this result more intuitive under a symmetric formulation of duality.

Given a scalar product (or pairing) between two vector spaces $V^*$ and $V$ over a field $k$ -- that is, a non-degenerate bilinear form $\langle-,-\rangle:V^*\times V\to k$ -- the result says that for any subspace $W$ of $V$ there is an induced scalar product between $V^*/W^{\perp}$ and $W$ given by $$\langle\overline{x^*},x\rangle=\langle x^*,x\rangle$$ where $W^{\perp}=\{\,x^*\in V^*\mid\langle x^*,x\rangle=0\text{ for all }x\in W\,\}$ is the orthogonal complement (or annihilator) of $W$. Symmetrically for any subspace $W^*$ of $V^*$ there is an induced scalar product between $W^*$ and $V/(W^*)^{\perp}$.

The principle here is that if you restrict to a subspace in one dual space, you need to take the quotient of its orthogonal complement in the other in order to preserve duality (non-degeneracy).

Intuitively the result tells us that subspaces are dual to quotient spaces. Statements about one can be translated into statements about the other, although sometimes a statement is more natural for one versus the other.

blargoner
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