Definition Let \(V\) be a vector space over a field \(K\). Then the dual space of \(V\) is \(V^*=\mathcal{L}(V,K)\). That is, the space of linear functions or functionals on \(V\).

If \(\{e_i\}\) is a basis of an \(n\)-dimensional vector space \(V\), then we can define elements \(e^i\) in \(V^*\) according to \(e^i(e_j)=\delta^i_j\) with the obvious linear extension to the whole of \(V\). It is not difficult to see that the \(e^i\) form a basis of \(V^*\). Indeed if \(c_ie^i=0\) then \(c_ie^i(e_j)=c_j=0\) for all \(j\) and if \(\varphi\in V^*\) then we can write \(\varphi=\varphi(e_i)e^i\) since on any basis element, \(e_j\), \(\varphi(e_i)e^i(e_j)=\varphi(e_j)\). Given a basis \(\{e_i\}\) of \(V\), the basis \(\{e^i\}\) so defined is called the dual basis of \(V^*\).

As \(\dim V^*=\dim V\) we have \(V^*\cong V\). Note that this isomorphism is not basis independent. In other words it is not canonical. By contrast, the vector spaces \(V\) and \(V^{**}\) are canonically isomorphic. Indeed, consider the map \(V\mapto V^{**}\) which takes \(v\mapsto\tilde{v}\) where \(\tilde{v}(\varphi)=\varphi(v)\) for any \(\varphi\in V^*\). Then it is not difficult to see that \(\tilde{v}\) so defined is certainly a linear function on \(V^*\) and that the map \(v\mapsto\tilde{v}\) is linear. It is injective since if \(v\neq0\) then we can extend \(v\) to a basis of \(V\) and so define on \(V\) a linear function \(\varphi\) such that \(\varphi(v)\neq0\) so \(\tilde{v}(\varphi)\neq0\) and thus \(\tilde{v}\neq0\). Since \(V\) is finite dimensional the map is an isomorphism.

Given a choice of basis, vectors \(v\) in \(V\) were identified with the column vectors, \(\mathbf{v}\), of their components. Similarly, given a choice of basis of the dual space, linear functionals \(\varphi\in V^*\) can naturally be identified with row vectors \(\boldsymbol{\varphi}\) of their components. Then \(\varphi(v)\) is simply the matrix product of the \(1\times n\) row vector \(\boldsymbol{\varphi}\) with the \(n\times 1\) column vector \(\mathbf{v}\), that is, \(\varphi(v)=\boldsymbol{\varphi}\mathbf{v}\).

We’ve seen that when we make a change of basis in \(V\), with \(e’_i=P_i^je_j\), then \(\mathbf{v}’=\mathbf{P}^{-1}\mathbf{v}\). The corresponding dual bases, \(\{e^i\}\) and \(\{e’^j\}\), must be related as say, \(e’^i=Q^i_je^j\), where \(Q^i_j\) are elements of an invertible matrix \(\mathbf{Q}\). Then we have \(e’^i(e’_j)=\delta^i_j=(Q^ike^k)(P^l_je_l)=Q^i_kP^k_j\). That is, \(\mathbf{Q}=\mathbf{P}^{-1}\). Now for any \(\varphi\in V^*\) we have the alternative expansions \(\varphi=\varphi_ie^i=\varphi’_ie’^i\), from which it follows that \(\boldsymbol{\varphi}’=\boldsymbol{\varphi}\mathbf{P}\). The components of vectors in the dual space thus transform with the change of basis matrix and so are said to transform covariantly.

Suppose \(V\) and \(W\) are respectively \(n\) and \(m\)-dimensional vector spaces over \(K\), with \(T\in\mathcal{L}(V,W)\). Then we can define the dual map \(T^*\), \(T^*:W^*\mapto V^*\) as \(T^*\omega=\omega T\). Having chosen bases for \(V\) and \(W\), \(\{e_i\}\) and \(\{f_i\}\) respectively, so that with respect to these bases \(\mathbf{T}\) is the matrix representation of \(T\), let us consider the matrix representation of \(T^*\) with respect to the dual bases \(\{e^i\}\) and \(\{f^i\}\). We have,
\begin{equation*}
(T^*f^i)(e_j)=f^i(Te_j)=f^i(T_j^kf_k)=T_j^i
\end{equation*}
but also
\begin{equation*}
(T^*f^i)(e_j)=((T^*)^i_ke^k)(e_j)=(T^*)^i_j,
\end{equation*}
that is \(\mathbf{T}^*=\mathbf{T}\). Note that if we’d chosen to identify elements of the dual space with column vectors of components then we would have found that the matrix of the dual map with respect to the dual bases was the transpose of the matrix of the original map.

Definition If \(U\) is a subspace of \(V\) then the annihilator of \(U\), \(U^\circ\) is defined to be
\begin{equation}
U^\circ=\{\varphi\in V^*\mid\varphi(u)=0\;\forall u\in U\}.
\end{equation}

Given a subspace \(U\) of \(V\) any element \(\varphi\in V^*\) has a restriction \(\varphi|_U\). This defines a linear operator \(\pi_U:V^*\mapto U^*\) as \(\pi_U(\varphi)=\varphi|_U\). Notice that \(\ker\pi_U=U^\circ\) and also that \(\img\pi_U=U^*\) so, since \(\dim V=\dim V^*\), we have
\begin{equation}
\dim U+\dim U^\circ=\dim V.\label{equ:annihilator dimension}
\end{equation}

Now suppose \(T\in\mathcal{L}(V,W)\), so that \(T^*\in\mathcal{L}(W^*,V^*)\), and consider \(\ker T^*\). That is, we consider elements \(\omega\in W^*\) such that \(T^*\omega=\omega T=0\). It is not difficult to see that this is precisely the annihilator of \(\img T\), that is
\begin{equation}
\ker T^*=(\img T)^\circ.
\end{equation}
Having observed that they have the same matrix representation (albeit modulo a different convention with regard to row and column vectors), we know that \(T\) and \(T^*\) have the same rank. But we can see this in a basis free fashion as follows. By the rank-nullity theorem we have \(\dim\img T^*=\dim W-\dim\ker T^*\). That is, \(\dim\img T^*=\dim W-\dim(\img T)^\circ\), but by \eqref{equ:annihilator dimension}, this just says that
\begin{equation}
\dim\img T^*=\dim\img T.
\end{equation}
Finally, maintaining a certain symmetry, we have
\begin{equation}
\img T^*=(\ker T)^\circ.
\end{equation}
To see this, consider \(\varphi\in\img T^*\), that is \(\varphi=\omega T\) for some \(\omega\in W^*\). Now if \(v\in\ker T\) then clearly \(\varphi(v)=\omega(Tv)=0\) so \(\varphi\in(\ker T)^\circ\). Thus we certainly have \(\img T^*\subseteq(\ker T)^\circ\). But \(\dim\ker T=\dim V-\dim\img T\) from the rank-nullity theorem, and we have just observed that \(\dim\ker T=\dim V-\dim(\ker T)^\circ\) so indeed \(\dim\img T=\dim(\ker T)^\circ\).