Intuition for why the dual of the quotient space $(V/W)’$ is isomorphic to the annihilator $W^0$

dual-spaceslinear algebralinear-transformations

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.

Best Answer

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.

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