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).)
