91Ó°ÊÓ

A linear operator A on a finite-dimensional inner-product space V is called self-adjoint if for all vectors u, v in V, the following condition holds: \(\langle Au, v \rangle = \langle u, Av \rangle\). In other words, A is self-adjoint if it equals its adjoint, which is denoted as \(A = A^*\).

Let A and B be two self-adjoint operators on the same finite-dimensional inner-product space V. Define the product AB of these operators as the result of applying operator B first to the vector, followed by applying operator A to the result. That is, for any vector u in V, we have \((AB)u = A(Bu)\).

In order to determine whether the product of two self-adjoint operators is self-adjoint, we need to check if \(\langle (AB)u, v \rangle = \langle u, (AB)v \rangle\) holds for all vectors u, v in V. Let's check this relation: \[\begin{aligned} \langle (AB)u, v \rangle &= \langle A(Bu), v \rangle \\ &= \langle Bu, A^*v \rangle &\text{(Since A is self-adjoint, }A = A^*) \\ &= \langle Bu, Av \rangle \\ &= \langle u, B^*Av \rangle &\text{(Since B is self-adjoint, }B = B^*) \\ \end{aligned}\] At this point, we can't automatically claim that the inner product \(\langle u, B^*Av \rangle\) is equal to \(\langle u, (AB)v \rangle\). To see this, let's compute \(\langle u, (AB)v \rangle\): \[\begin{aligned} \langle u, (AB)v \rangle &= \langle u, A(Bv) \rangle \\ \end{aligned}\] Now, we can see that the two expressions are not directly equal, which means that the statement might not always hold. Therefore, we will consider an example to see if we can find a counterexample.

Let's consider a 2-dimensional real inner-product space and define two self-adjoint operators A and B as follows: \[A = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \quad B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\] Note that both A and B are self-adjoint as their transposes equal their (negative) inverse, i.e., \(A = A^T\) and \(B = B^T\). Now, let's compute the product AB and its adjoint: \[AB = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\] And, \[(AB)^T = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\] In this case, we can clearly see that \(AB \neq (AB)^T\), which means that the product of these two self-adjoint operators is not self-adjoint. Thus, we have found a counterexample. In conclusion, the statement is false: the product of any two self-adjoint operators on a finite-dimensional inner-product space is not always self-adjoint, as demonstrated by our counterexample.