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The dual space isthe vector space of linear functionals on $V$. If $e_1,\dots ,e_n$ is a basis of a finite-dimensional $V$, define $e_i^{\ast }(e_j)={\delta }_{ij}$. Every $f\in V^{\ast }$ has the unique expansionso the dual basis $e_1^{\ast },\dots ,e_n^{\ast }$ proves

For $U\le V$, its annihilator of a vector subspace isIf $u_1,\dots ,u_k$ is a basis of $U$ and is extended to a basis $u_1,\dots ,u_n$ of $V$, thenConsequentlyIf $U\ne V$, this dimension is positive, giving a nonzero functional that vanishes on $U$.

For a linear map $\alpha :V\to W$, the dual map isNowsoEvery $g\circ \alpha$ vanishes on $\ker \alpha$, henceThe two spaces have the same dimension, sinceThus

Let $Q:V\to V/U$ be the quotient map. Since $Q$ is onto, $Q^{\ast }$ is injective, and the preceding identity givesThereforeFor the inclusion $J:U\hookrightarrow V$, the map $J^{\ast }:V^{\ast }\to U^{\ast }$ is restriction to $U$. It is onto and has kernelThe first isomorphism theorem now gives the other duals of a subspace and its quotient: