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The algebraic double dual of a vector space is the dual space of its algebraic dual. An element assigns a scalar linearly to each linear functional on the original space. The bidual evaluation map sends $v$ to the rule $\ell \mapsto \ell (v)$. It is a canonical linear isomorphism in finite dimension, unlike an identification of $V$ with $V^{\ast }$, which generally requires an additional choice. For spaces with topology, using the continuous dual space instead defines the topological bidual; that is a distinct construction in infinite dimension.

For a bilinear form $B$ on a finite-dimensional vector space $V$, let $\Phi :V\to V^{\ast }$ send $u$ to $B(u,\cdot )$. Under the bidual evaluation map, $U^{\perp }$ is the annihilator of a vector subspace $\mathrm{im}(\Phi {|}_U)$, so $\dim U^{\perp }=\dim V-\mathrm{rank}(\Phi {|}_U)$. The rank-nullity theorem on $U$ now gives the displayed equality. In particular the sum of dimensions is at least $\dim V$. If $U\cap U^{\perp }=0$, the dimension formula for a sum of subspaces forces $V=U\oplus U^{\perp }$. The exact formula records the extra dimensions contributed by the left radical $\ker \Phi$, and does not assume an inner product.

Under the bidual evaluation map, a vector $y$ lies in $(\mathrm{im}\Phi {|}_U{)}^{\circ }$ exactly whenThat is precisely membership of the orthogonal complement for a bilinear form. ThereforeThe dimension formula in part (ii) givesso $\boxed{\dim U+\dim U^{\perp }\ge n}$. More precisely, the rank-nullity theorem givesThis is the orthogonal complement dimension for a degenerate bilinear form. The possible excess comes from vectors annihilated by the form against all of $V$.

The annihilator of a vector subspace $W\le V^{\ast }$ isIf $\dim W=k$, extend a basis of $W$ to one of $V^{\ast }$. In the resulting dual basis of $V^{\ast \ast }$, the last $n-k$ coordinate functionals span $W^{\circ }$. Hence $\boxed{\dim W+\dim W^{\circ }=n}$.

The bidual evaluation map $j:V\to V^{\ast \ast }$ is $j(x)(\varphi )=\varphi (x)$. It is a linear isomorphism in finite dimension. Its inverse image of the annihilator isThis identification uses evaluation only; no choice of inner product is involved.

Define the bidual evaluation map by evaluation:For fixed $v$, this is a linear functional of $\ell$, so it belongs to the double dual. It is also linear as a function of $v$. If $v\ne 0$, some coordinate of $v$ in a basis is nonzero, and the corresponding dual basis functional detects it. Hence $J(v)\ne 0$, proving injectivity.

The spaces $V,V^{\prime },V^{\prime \prime }$ all have the same finite dimension. Therefore an injective linear map $J:V\to V^{\prime \prime }$ is surjective, proving the required linear isomorphism.

This linear isomorphism is natural because evaluation requires no basis. More explicitly, for any linear map $A:V\to W$, its dual map satisfiesso $A^{\prime \prime }{J}_V=J_{W}A$.

We prove by mathematical induction on $n=\dim V$ that every linear map on an $n$-dimensional complex vector space admits both triangular forms. For $n=0$ or $1$ this is immediate. When $n>1$, the fundamental theorem of algebra supplies an eigenvector $v$; its span $U=⟨v⟩$ is an invariant subspace.

By parts (ii) and (iii), $U^0$ has dimension $n-1$ and is an invariant subspace of the dual map $T^{\prime }$. Apply the induction hypothesis to $T^{\prime }{|}_{U^0}$ to choose a basis ${\phi }_1,\dots ,{\phi }_{n-1}$ with an upper triangular matrix. Extend it by a linear functional ${\phi }_n$ satisfying ${\phi }_n(v)=1$. Such a functional exists by basis extension, and cannot belong to $U^0$. In this basis of $V^{\prime }$, the matrix of $T^{\prime }$ has the formwhere $B$ is an upper triangular matrix; here the last diagonal entry follows by evaluating $T^{\prime }{\phi }_n$ at $v$, where $Tv=\lambda v$.

There is a basis $e_1,\dots ,e_n$ of $V$ with ${\phi }_i(e_j)={\delta }_{ij}$, since the dual basis construction gives a finite-dimensional bidual evaluation map isomorphism. In particular $e_n=v$. The matrix transpose relationship for a dual map givesa lower triangular matrix. Reversing this basis turns a lower triangular matrix into an upper triangular matrix. If $R$ is the permutation matrix reversing the order, we have provedThis completes the induction by triangularization using dual spaces.

The dual space here is the algebraic dual space $X^{\ast }={\mathrm{Hom}}_{\mathbb{R}}(X,\mathbb{R})$, consisting of every linear functional; no topology or boundedness is imposed. For a basis $e_1,\dots ,e_n$, define $e_i^{\ast }({\sum }_j{x}_j{e}_j)=x_i$. These are linear functionals satisfying $e_i^{\ast }(e_j)={\delta }_{ij}$. Evaluating ${\sum }_i{a}_i{e}_i^{\ast }=0$ on each $e_j$ gives $a_j=0$, proving linear independence. Conversely, for every $T\in X^{\ast }$,This proves spanning and hence proves, without assuming a dimension theorem, that the dual basis is a basis and $\dim X^{\ast }=n$.

The canonical bidual evaluation map isIts definition makes no choice of basis and it is linear. If $x\ne 0$, one coordinate functional is nonzero on $x$, so $J$ is injective. For any $B\in X^{\ast \ast }$, choose $x={\sum }_{i}B(e_i^{\ast })e_i$. Expanding $T$ in the dual basis gives $T(x)={\sum }_{i}T(e_i)B(e_i^{\ast })=B(T)$, proving surjection explicitly. Thus $J$ is an isomorphism.

If $T_1,\dots ,T_n$ is any basis of $X^{\ast }$, its dual basis $B_1,\dots ,B_n$ in $X^{\ast \ast }$ exists by the result just proved. Put $v_j=J^{-1}(B_j)$. These vectors form a basis of $X$, and $T_i(v_j)=B_j(T_i)={\delta }_{ij}$. Every basis of the finite-dimensional dual is a dual basis, of a uniquely determined basis of $X$.

For a separating subspace of an algebraic dual $W$, choose a basis $T_1,\dots ,T_r$ of $W$. The linear map $x\mapsto (T_1(x),\dots ,T_r(x))$ is injective because $W$ separates nonzero vectors. The rank-nullity theorem implies $n\le r$; but $W\subseteq X^{\ast }$ implies $r\le n$. HenceThe zero-dimensional case also has this conclusion.

For $X=\mathbb{R}[x]$, the monomials $1,x,x^2,\dots$ form an infinite basis, each polynomial using only finitely many of them. The map $T\mapsto (T(1),T(x),T(x^2),\dots )$ identifies $X^{\ast }$ with all real sequences, because any sequence $(a_j)$ defines $T({\sum }_{j=0}^d{c}_j{x}^j)={\sum }_{j=0}^d{c}_j{a}_j$. The linear subspace of finitely supported sequences is proper (it excludes $(1,1,\dots )$) and separating (one coefficient functional detects any nonzero polynomial). Thus