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For a vector subspace $U\le V$ of a finite-dimensional vector space, there are canonical linear isomorphismswhere $U^{\circ }$ is the annihilator of a vector subspace.

For the first identity, let $q:V\to V/U$ be the quotient map of vector spaces. Composing a linear functional on $V/U$ with $q$ gives a linear functional on $V$ that vanishes on $U$. Conversely, if $f\in U^{\circ }$, the formula $\bar{f}(v+U)=f(v)$ is well-defined: changing the representative adds an element on which $f$ is zero. These operations are inverse.

For the second identity, restriction $V^{\ast }\to U^{\ast }$ is surjective by basis extension, and its kernel is exactly $U^{\circ }$. Two linear functionals therefore have the same restriction precisely when their difference belongs to this annihilator of a vector subspace. Taking the quotient vector space gives the identity.

For example, if $V={\mathbb{R}}^3$ and $U$ is the span of $(1,0,0)$, then $U^{\circ }$ consists of $f(x,y,z)=by+cz$. These linear functionals descend to $V/U$. Restricting a general $f(x,y,z)=ax+by+cz$ to $U$ instead remembers only $a$; the terms in $y,z$ disappear in the quotient of the dual.

The quotient map of vector spaces $\pi :V\to V/U$ sends a vector $v$ to its coset $v+U$, where $U$ is a linear subspace. It is a surjective linear map with kernel $U$: two vectors have the same image exactly when their difference lies in $U$.

For $V={\mathbb{R}}^2$ and $U=\mathrm{span}(1,0)$, the quotient retains only the second coordinate, so every horizontal line represents one quotient vector. More generally, a linear map $f:V\to W$ vanishing on $U$ factors uniquely as $f=\overline{f}\circ \pi$. The formula $\overline{f}(v+U)=f(v)$ is well defined because changing the representative by an element of $U$ leaves its value unchanged.