Can an operator not be self-adjoint even though its representation is?

Question

Let $V = P_2(\mathbb R)$ be the space of all polynomials of degree not exceeding 2, endowed with the inner product $\langle f(t),g(t) \rangle$ = $\int_0^1 f(t)g(t)dt$. Suppose that $T : V → V$ is the linear map defined by $T(a_0 + a_1 t + a_2 t^2 ) = a_1 t$.

(a) Demonstrate that $T$ is not self-adjoint by considering $\langle T(1),t \rangle$.

(b) Let $C = \{1,t,t^2\} $ and $B = [T]_C$ . Show that $B^∗ = B$.

(c) Although $B^∗ = B$, $T$ is not self-adjoint as observed in part (a). Why is this not a contradiction? Justify.

I got the first and second parts in this. The third part however, I'm not sure if I got the right answer. My reasoning was that if the operator $T$ is self-adjoint, then its representations are hermitian, and the converse need not be true. Any suggestions to make this reasoning more rigorous would be really helpful.

Answer 1

The equivalence

$T$ is self-adjoint $\Longleftrightarrow$ $[T]_C$ is a Hermitian matrix

is only true when $C$ is an orthonormal basis. If $C$ is any basis, neither $\Longrightarrow$ nor $\Longleftarrow$ need to hold.