Let us study next the relation between the notions of linear transformation and dual space. Let \(\mathcal{V}\) be any vector space and let \(y\) be any element of \(\mathcal{V}^{\prime}\) ; for any linear transformation \(A\) on \(\mathcal{V}\) we consider the expression \([A x, y]\) . For each fixed \(y\) , the function \(y^{\prime}\) defined by \(y^{\prime}(x)=[A x, y]\) is a linear functional on \(\mathcal{V}\) ; using the square bracket notation for \(y^{\prime}\) as well as for \(y\) , we have \([A x, y]=[x, y^{\prime}]\) . If now we allow \(y\) to vary over \(\mathcal{V}^{\prime}\) , then this procedure makes correspond to each \(y\) a \(y^{\prime}\) , depending, of course, on \(y\) ; we write \(y^{\prime}=A^{\prime} y\) . The defining property of \(A^{\prime}\) is \[=[x, A^{\prime} y]. \tag{1}\] We assert that \(A^{\prime}\) is a linear transformation on \(\mathcal{V}^{\prime}\) . Indeed, if \(y=\alpha_{1} y_{1}+\alpha_{2} y_{2}\) , then \begin{align} [x, A^{\prime} y] &= [A x, y]\\ &= \alpha_{1}[A x, y_{1}]+\alpha_{2}[A x, y_{2}] \\ &= \alpha_{1}[x, A^{\prime} y_{1}]+\alpha_{2}[x, A^{\prime} y_{2}]\\ &=[x, \alpha_{1} A^{\prime} y_{1}+\alpha_{2} A^{\prime} y_{2}]. \end{align} The linear transformation \(A^{\prime}\) is called the adjoint (or dual) of \(A\) ; we dedicate this section and the next to studying properties of \(A^{\prime}\) . Let us first get the formal algebraic rules out of the way; they go as follows. \begin{align} 0^{\prime} & =0, \tag{2}\\ 1^{\prime} & =1, \tag{3}\\ (A+B)^{\prime} & =A^{\prime}+B^{\prime}, \tag{4}\\ (\alpha A)^{\prime} & =\alpha A^{\prime}, \tag{5}\\ (A B)^{\prime} & =B^{\prime} A^{\prime}, \tag{6}\\ (A^{-1})^{\prime} & =(A^{\prime})^{-1}. \tag{7} \end{align} 

Here (7) is to be interpreted in the following sense: if \(A\) is invertible, then so is \(A^{\prime}\) , and the equation is valid. The proofs of all these relations are elementary; to indicate the procedure, we carry out the computations for (6) and (7). To prove (6), merely observe that \[[A B x, y]=[B x, A^{\prime} y]=[x, B^{\prime} A^{\prime} y].\] To prove (7), suppose that \(A\) is invertible, so that \(A A^{-1}=A^{-1} A=1\) . Applying (3) and (6) to these equations, we obtain \[(A^{-1})^{\prime} A^{\prime}=A^{\prime}(A^{-1})^{\prime}=1;\] Theorem 1 of Section: Inverses implies that \(A^{\prime}\) is invertible and that (7) is valid.

In finite-dimensional spaces another important relation holds: \[A^{\prime \prime}=A. \tag{8}\] This relation has to be read with a grain of salt. As it stands \(A^{\prime \prime}\) is a transformation not on \(\mathcal{V}\) but on the dual space \(\mathcal{V}^{\prime \prime}\) of \(\mathcal{V}^{\prime}\) . If, however, we identify \(\mathcal{V}^{\prime \prime}\) and \(\mathcal{V}\) according to the natural isomorphism, then \(A^{\prime \prime}\) acts on \(\mathcal{V}\) and (8) makes sense. In this interpretation the proof of (8) is trivial. Since \(\mathcal{V}\) is reflexive, we obtain every linear functional on \(\mathcal{V}^{\prime}\) by considering \([x, y]\) as a function of \(y\) , with \(x\) fixed in \(\mathcal{V}\) . Since \([x, A^{\prime} y]\) defines a function (a linear functional) of \(y\) , it may be written in the form \([x^{\prime}, y]\) . The vector \(x^{\prime}\) here is, by definition, \(A^{\prime \prime} x\) . Hence we have, for every \(y\) in \(\mathcal{V}^{\prime}\) and for every \(x\) in \(\mathcal{V}\) , \[[A x, y]=[x, A^{\prime} y]=[A^{\prime \prime} x, y];\] the equality of the first and last terms of this chain proves (8).

Under the hypothesis of (8) (that is, finite-dimensionality), the asymmetry in the interpretation of (7) may be removed; we assert that in this case the invertibility of \(A^{\prime}\) implies that of \(A\) and, therefore, the validity of (7). Proof: apply the old interpretation of (7) to \(A^{\prime}\) and \(A^{\prime \prime}\) in place of \(A\) and \(A^{\prime}\) .

Our discussion is summed up, in the reflexive finite-dimensional case, by the assertion that the mapping \(A \to A^{\prime}\) is one-to-one, and, in fact, an algebraic anti-isomorphism, from the set of all linear transformations on \(\mathcal{V}\) onto the set of all linear transformations on \(\mathcal{V}^\prime\) . (The prefix "anti" got attached because of the commutation rule (6).)