91Ó°ÊÓ

- Inner-product space: An inner-product space V is a vector space equipped with an inner product, which is a function that assigns a scalar to each pair of vectors. - Self-adjoint operator: An operator \(N \in \mathcal{L}(V)\) is self-adjoint if \(\langle Nv, w \rangle = \langle v, Nw \rangle\) for all \(v, w \in V\). - Nilpotent operator: An operator N is nilpotent if there exists a positive integer k such that \(N^k = 0\), where \(N^k\) denotes applying the operator N k times.

We need to show that if N is both self-adjoint and nilpotent, then N must be the zero operator. Let's use the given information: Since N is self-adjoint, we know that \(\langle Nv, w \rangle = \langle v, Nw \rangle\) for all \(v, w \in V\). Since N is nilpotent, there exists a positive integer k such that \(N^k = 0\). Now let's take any vector \(z = Nv \in V\), for some \(v \in V\). Then \(\langle Nz, w \rangle = \langle N^2v, w \rangle = \langle v, NNw \rangle = \langle v, N^2w \rangle\). Now let's calculate \(\langle z, z \rangle\): \(\langle z, z \rangle = \langle Nz, Nz \rangle\) (as z = Nv) \(\langle z, z \rangle = \langle N^2v, N^2v \rangle\) (since Nz = N^2v) \(\langle z, z \rangle = \langle v, N^4v \rangle\) (same self-adjoint rule as before) Keep repeating this process until: \(\langle z, z \rangle = \langle v, N^{2k-2}v \rangle\) But we know that \(N^k = 0\), hence \(N^{2k-2}\) is also 0 because \(2k-2 \ge k\). Therefore, \(\langle z, z \rangle = \langle v, 0 \rangle = 0\) Now we know that an inner product \(\langle u, u \rangle = 0\) if and only if u = 0. In our case, \(\langle z, z \rangle = 0 \Rightarrow z = 0\), thus \(Nv = 0\). Since Nv = 0 for any vector v in V, it follows that N must be the zero operator on V.

We have proven that if an operator N is self-adjoint and nilpotent on an inner-product space V, then N must be the zero operator.