Linear Algebra Done Right · Problem 7A.14

Operators on Inner Product Spaces — Sheldon Axler

Mathematics — Linear Algebra · self-adjoint-operators

Problem

Define an inner product on

\mathcal{P}_2(\mathbf{R})

\langle p, q \rangle = \int_0^1 pq

. Define an operator

T \in \mathcal{L}(\mathcal{P}_2(\mathbf{R}))

T(ax^2 + bx + c) = bx.

(a) Show that with this inner product, the operator

T

is not self-adjoint. (b) The matrix of

T

with respect to the basis

1, x, x^2

\begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}.

This matrix equals its conjugate transpose, even though

T

is not self-adjoint. Explain why this is not a contradiction.