Adjoints of projections

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 $ℳ$ along $𝒩$ , then E^{\prime} is the projection on $𝒩^{0}$ along $ℳ^{0}$ .

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 $ℳ^{0}$ , 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 $ℳ$ , [x, y]=[E x, y]=[x, E^{\prime} y]=0, so that $y$ is in $ℳ^{0}$ .
If $y$ is in $𝒩^{0}$ , 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 $𝒩$ , [x, y]=[x, E^{\prime} y]=[E x, y]=0, so that $y$ is in $𝒩^{0}$ .

Steps (i) and (ii) together show that the set of solutions of E^{\prime} y=0 is precisely $ℳ^{0}$ ; steps (iii) and (iv) together show that the set of solutions of E^{\prime} y=y is precisely $𝒩^{0}$ . This concludes the proof of the theorem. ◻

Theorem 2. If $ℳ$ is invariant under $A$ , then $ℳ^{0}$ is invariant under A^{\prime} ; if $A$ is reduced by $(ℳ, 𝒩)$ , then A^{\prime} is reduced by $(ℳ^{0}, 𝒩^{0})$ .

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 $ℳ$ ; 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 $ℳ^{0}$ , 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 $𝒱$ is the direct sum of $ℳ$ and some other subspace. If $y$ is in $ℳ^{0}$ , then [x, A^{\prime} y]=[A x, y]=0 for all $x$ in $ℳ$ , and therefore A^{\prime} y is in $ℳ^{0}$ . 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 $𝒳 = {x_{1}, \dots, x_{n}}$ is any basis in the $n$ -dimensional vector space $𝒱$ , 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 $𝒳$ is $(α_{i j})$ , then \alpha_{i j}=[A x_{j}, y_{i}]. \tag{2} This follows from the definition of the matrix of a linear transformation; since $A x_{j} = \sum_{k} α_{k j} x_{k}$ , we have $[A x_{j}, y_{i}] = \sum_{k} α_{k j} [x_{k}, y_{i}] = α_{i j} .$ To keep things straight in the applications, we rephrase formula (2) verbally, thus: to find the $(i, j)$ element of $[A]$ in the basis $𝒳$ , apply $A$ to the $j$ -th element of $𝒳$ 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.

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