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For a linear map between finite-dimensional vector spaces, the image of its dual map is the annihilator of a vector subspace given by its kernel. Inclusion follows by composition. Conversely a linear functional vanishing on the kernel descends to the image and extends to the target by basis extension. The identity and the rank-nullity theorem show that a linear map and its dual have equal rank.

For a linear map $T:V\to W$ with $V$ finite-dimensional, the rank-nullity theorem statesTo prove it, choose a basis $v_1,\dots ,v_k$ of the kernel of a linear map and use basis extension to obtain $v_1,\dots ,v_n$ as a basis of $V$. The vectors $T{v}_{k+1},\dots ,T{v}_n$ span the image of a linear map: applying $T$ to a basis expansion kills the first $k$ terms. They are also linearly independent. Indeed, if ${\sum }_{j>k}{a}_{j}T{v}_j=0$, then ${\sum }_{j>k}{a}_j{v}_j\in \ker T$, so it is a combination of $v_1,\dots ,v_k$. Independence of the full basis forces every $a_j=0$. Consequently $\dim \mathrm{im}T=n-k$, proving the theorem.

Here the image of a linear map has dimension $r\in \{0,1,2,3\}$, since it lies in ${\mathbb{R}}^3$. ThusEvery possibility occurs: for $0\le r\le 3$, map $(x_1,\dots ,x_5)$ to $(x_1,\dots ,x_r,0,\dots ,0)\in {\mathbb{R}}^3$. This map has rank $r$ and kernel dimension $5-r$.

The algebraic dual map is $\delta ={\alpha }^{\vee }:W^{\ast }\to V^{\ast }$, defined by $\delta (\phi )=\phi \circ \alpha$. The annihilator of a vector subspace $U\subseteq V$ isEvery functional $\phi \circ \alpha$ vanishes on $\ker \alpha$, so $\mathrm{im}\delta \subseteq (\ker \alpha {)}^{\circ }$. Conversely, let $\lambda \in (\ker \alpha {)}^{\circ }$. Define a functional on $\mathrm{im}\alpha$ by $\tilde{\lambda }(\alpha v)=\lambda (v)$. It is well defined because two preimages differ by an element of $\ker \alpha$. Extend a basis of $\mathrm{im}\alpha$ to one of $W$ and extend the functional arbitrarily, for instance by zero on the extra basis vectors. The resulting $\phi \in W^{\ast }$ satisfies $\delta \phi =\lambda$.

This is the dual image and kernel annihilator identity. ThereforeThe basis extension theorem also gives $\dim U^{\circ }=\dim V-\dim U$. By the rank-nullity theorem,Apply rank-nullity theorem to both maps, using $\dim W^{\ast }=\dim W$, to obtainOver $\mathbb{C}$, this is the algebraic dual, without complex conjugation; the adjoint operator in part (b) uses the inner products instead.