Doubt about property of hermitian operator

Question

For any hermitian operator M, prove that \begin{equation} \langle Ma|b \rangle = \langle a|Mb \rangle \end{equation}

My attempt:
Let \begin{eqnarray} \langle a| = \sum_i a_i^*\langle i|\\ |b\rangle = \sum_j b_j|j\rangle \end{eqnarray} Then \begin{eqnarray} \langle Ma| &=& \sum_{i} a_i^*M\langle i|\\ \therefore\langle Ma|b\rangle &=& \sum_{i,j} a_i^*b_jM\langle i|j\rangle\\ &=& \sum_{i,j} a_i^*b_jM\delta_{ij}\\ &=&\sum_{i} a_i^*b_iM \end{eqnarray} Now similarly, \begin{eqnarray} \langle a|Mb\rangle &=& \sum_{i,j} a_i^*b_j\langle i|M|j \rangle\\ &=& \sum_{i,j} a_i^*b_j M_{ij} \end{eqnarray} I don't know what to do next. I feel that I am very close to the answer

Answer 1

As pointed out in the comment, The identity that you are trying to prove is mainly taken as the definition of a Hermitian operator.

A is Hermitian if $A=A^\dagger$ or equivalently if $\langle x,Ay\rangle =\langle Ax,y\rangle \ \ \forall\ \ x,y.$

This can be seen as follows: $$\langle a|A^\dagger|b\rangle \equiv \langle b|A|a\rangle ^*$$ At this point, We can proceed in two equivalent way $$\langle a|A^\dagger |b\rangle=\langle b|aA\rangle^*=\langle Aa|b\rangle$$ Or $$\langle a|A^\dagger|b\rangle =\langle Ab|a\rangle^*=\langle a|bA\rangle $$ Hence$$\langle a|bA\rangle =\langle Aa|b\rangle $$