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

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.

• Because taking the adjoint depends on the scalar product and we consider two different scalar products. Dec 13, 2020 at 16:28
• Two different scalar products? And by scalar product you mean the inner product right? Dec 13, 2020 at 16:31
• Indeed, that is what I mean Dec 13, 2020 at 17:09
• Okay, but I don't understand what you mean by "two different scalar products" Dec 13, 2020 at 17:13
• Taking the hermitean conjugate of $B$ is the same as taking the adjoint with respect to the standard scalar product. However, on $V$ you have another scalar product. Dec 13, 2020 at 17:27

$$T$$ 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.