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For an algebraic dual space $X^{\ast }={\mathrm{Hom}}_F(X,F)$, the linear map $J:X\to X^{\ast \ast }$ sends $x$ to evaluation at $x$. It is injective because linear functionals separate nonzero vectors: extend any nonzero vector to a basis and use its coordinate functional. If $X$ has a finite basis $e_i$ with dual basis $e_i^{\ast }$, every $B\in X^{\ast \ast }$ equals $J({\sum }_{i}B(e_i^{\ast })e_i)$, so $J$ is an isomorphism. For infinite-dimensional spaces, surjection does not follow from this finite sum argument.

The coadjoint action of a Lie group on the dual space of its Lie algebra is $({\mathrm{Ad}}_g^{\ast }\eta )(X)=\eta ({\mathrm{Ad}}_{g^{-1}}X)$. The inverse in this formula makes it a left group action.

The continuous dual $X^{\ast }$ of a topological vector space is the vector space of its continuous scalar-valued linear functionals. For a normed space these are exactly the bounded linear functionals.

A covector is an element of the dual space, hence a linear functional on a vector space.

The dual of a Lie algebra representation on $V$ acts on its dual space by $(x\varphi )(v)=-\varphi (xv)$. The minus sign makes this a Lie algebra homomorphism: taking transposes reverses the order in a commutator. It is the infinitesimal version of the dual representation of a group.

The dual space ${\mathfrak{g}}^{\ast }$ of a finite-dimensional Lie algebra has this canonical Poisson bracket. For a basis $v^i$, linear coordinates $x^i(\ell )=\ell (v^i)$ give $\{x^i,x^j\}=c_k^{ij}{x}^k$. The Lie algebra's Jacobi identity is precisely the coordinate Jacobi condition for a Poisson bivector. This construction is intrinsic on the dual space; writing the same formula in an arbitrary manifold chart does not by itself provide a compatible global atlas.

A bilinear form $B$ on a finite-dimensional vector space is nondegenerate whenequivalently when $w\mapsto B(-,w)$ is an isomorphism from $V$ to its dual space.