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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.

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    $\begingroup$ Because taking the adjoint depends on the scalar product and we consider two different scalar products. $\endgroup$ Dec 13, 2020 at 16:28
  • $\begingroup$ Two different scalar products? And by scalar product you mean the inner product right? $\endgroup$ Dec 13, 2020 at 16:31
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    $\begingroup$ Indeed, that is what I mean $\endgroup$ Dec 13, 2020 at 17:09
  • $\begingroup$ Okay, but I don't understand what you mean by "two different scalar products" $\endgroup$ Dec 13, 2020 at 17:13
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    $\begingroup$ 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. $\endgroup$ Dec 13, 2020 at 17:27

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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.

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