There is one important case in which multiplication does not get turned around, that is, when (A B)^{\prime}=A^{\prime} B^{\prime} ; namely, the case when A and B commute. We have, in particular, (A^{n})^{\prime}=(A^{\prime})^{n} , and, more generally, (p(A))^{\prime}=p(A^{\prime}) for every polynomial p . It follows from this that if E is a projection, then so is E^{\prime} . The question arises: what direct sum decomposition is E^{\prime} associated with?
Theorem 1. If E is the projection on \mathcal{M} along \mathcal{N} , then E^{\prime} is the projection on \mathcal{N}^{0} along \mathcal{M}^{0} .
Proof. We know already that (E^{\prime})^{2}=E^{\prime} and \mathcal{V}^{\prime}=\mathcal{N}^{0} \oplus \mathcal{M}^{0} (cf. Section: Dual of a direct sum ). It is necessary only to find the subspaces consisting of the solutions of E^{\prime} y=0 and E^{\prime} y=y . This we do in four steps.
- If y is in \mathcal{M}^{0} , then, for all x , [x, E^{\prime} y]=[E x, y]=0, so that E^{\prime} y=0 .
- If E^{\prime} y=0 , then, for all x in \mathcal{M} , [x, y]=[E x, y]=[x, E^{\prime} y]=0, so that y is in \mathcal{M}^{0} .
- If y is in \mathcal{N}^{0} , then, for all x , \begin{align} &= [E x, y]+[(1-E) x, y]\\ &= [E x, y]\\ &= [x, E^{\prime} y], \end{align}so that E^{\prime} y=y .
- If E^{\prime} y=y , then for all x in \mathcal{N} , [x, y]=[x, E^{\prime} y]=[E x, y]=0, so that y is in \mathcal{N}^{0} .
Steps (i) and (ii) together show that the set of solutions of E^{\prime} y=0 is precisely \mathcal{M}^{0} ; steps (iii) and (iv) together show that the set of solutions of E^{\prime} y=y is precisely \mathcal{N}^{0} . This concludes the proof of the theorem. ◻
Theorem 2. If \mathcal{M} is invariant under A , then \mathcal{M}^{0} is invariant under A^{\prime} ; if A is reduced by (\mathcal{M}, \mathcal{N}) , then A^{\prime} is reduced by (\mathcal{M}^{0}, \mathcal{N}^{0}) .
Proof. We shall prove only the first statement; the second one clearly follows from it. We first observe the following identity, valid for any three linear transformations E , F , and A , subject to the relation F=1-E : F A F-F A=E A E-A E \tag{1} (Compare this with the proof of Section: Projections and invariance , Theorem 2.) Let E be any projection on \mathcal{M} ; by Section: Projections and invariance , Theorem 1, the right member of (1) vanishes, and, therefore, so does the left member. By taking adjoints, we obtain F^{\prime} A^{\prime} F^{\prime}=A^{\prime} F^{\prime} ; since, by Theorem 1 of the present section, F^{\prime}=1-E^{\prime} is a projection on \mathcal{M}^{0} , the proof of Theorem 2 is complete. (Here is an alternative proof of the first statement of Theorem 2, a proof that does not make use of the fact that \mathcal{V} is the direct sum of \mathcal{M} and some other subspace. If y is in \mathcal{M}^{0} , then [x, A^{\prime} y]=[A x, y]=0 for all x in \mathcal{M} , and therefore A^{\prime} y is in \mathcal{M}^{0} . The only advantage of the algebraic proof given above over this simple geometric proof is that the former prepares the ground for future work with projections.) ◻
We conclude our treatment of adjoints by discussing their matrices; this discussion is intended to illuminate the entire theory and to enable the reader to construct many examples.
We shall need the following fact: if \mathcal{X}=\{x_{1}, \ldots, x_{n}\} is any basis in the n -dimensional vector space \mathcal{V} , if \mathcal{X}^{\prime}=\{y_{1}, \ldots, y_{n}\} is the dual basis in \mathcal{V}^{\prime} , and if the matrix of the linear transformation A in the coordinate system \mathcal{X} is (\alpha_{i j}) , then \alpha_{i j}=[A x_{j}, y_{i}]. \tag{2} This follows from the definition of the matrix of a linear transformation; since A x_{j}=\sum_{k} \alpha_{k j} x_{k} , we have [A x_{j}, y_{i}]=\sum_{k} \alpha_{k j}[x_{k}, y_{i}]=\alpha_{i j}. To keep things straight in the applications, we rephrase formula (2) verbally, thus: to find the (i, j) element of [A] in the basis \mathcal{X} , apply A to the j -th element of \mathcal{X} and then take the value of the i -th linear functional (in \mathcal{X}^{\prime} ) at the vector so obtained.
It is now very easy to find the matrix (\alpha^{\prime}_{i j})=[A^{\prime}] in the coordinate system \mathcal{X}^{\prime} ; we merely follow the recipe just given. In other words, we consider A^{\prime} y_{j} , and take the value of the i -th linear functional in \mathcal{X}^{\prime \prime} (that is, of x_{i} considered as a linear functional on \mathcal{X}^{\prime} ) at this vector; the result is that \alpha_{i j}^{\prime}=[x_{i}, A^{\prime} y_{j}]. Since [x_{i}, A^{\prime} y_{j}]=[A x_{i}, y_{j}]=\alpha_{j i} , so that \alpha_{i j}^{\prime}=\alpha_{j i} , this matrix [A^{\prime}] is called the transpose of [A] .
Observe that our results on the relation between E and E^{\prime} (where E is a projection) could also have been derived by using the facts about the matricial representation of a projection together with the present result on the matrices of adjoint transformations.
