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For a linear functional $\lambda \in V^{\prime }$, the definition of the dual map givesThus the kernel of a linear map is precisely the annihilator of a vector subspaceWrite $n=\dim U$, $m=\dim V$ and $r=\dim \mathrm{im}T$. The rank-nullity theorem, the dimension formula for an annihilator of a vector subspace, and $\dim V^{\prime }=m$ giveApplying the same dimension formula in $U$ givesFinally, if $\alpha =T^{\prime }\lambda$ and $u\in \ker T$, then $\alpha (u)=\lambda (Tu)=\lambda (0)=0$. Therefore $\mathrm{im}{T}^{\prime }\subseteq (\ker T{)}^0$. These are finite-dimensional vector subspaces of equal dimension, so inclusion is equality:This is the dual image and kernel annihilator identity.

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.