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Definition

Let be $V$ and $U$ real vector spaces equipped with an inner product. So given a linear transformation $f:V\rightarrow U$ a function $f^*:U\rightarrow V$ is called the adjoint of $f$ if $$ \langle\vec u,f(\vec v)\rangle=\langle f^*(\vec u),\vec v\rangle $$ for all $\vec v\in V$ and for all $\vec u\in U$. In particular an endomorphism $f\in\mathscr L(V,V)$ is called Hermitian if $f=f^*$ and skew-Hermitian if $f=-f^*$.

Now let be $\mathscr B:=\{\vec e_1,...,\vec e_n\}$ an orthonormal basis for $V$. So we observe that $$ \langle f(\vec e_i),\vec e_j\rangle=\langle f^*(\vec e_i), \vec e_j\rangle=\langle \vec e_i,f(\vec e_j)\rangle $$ for each $i,j=1,...,n$ and we conclude that the matrix computed using the basis $\mathscr B$ is symmetric. So using the previous definition I ask if the matrix $A$ of an hermitian endomorphism is necessarly symmetric also if we don't compute the matrix using an orthonormal basis. Indeed generally if $\mathscr B$ is not orthomormal then the $a_{i,j}$ element of $A$ is given by the equation $$ a_{i,j}=\langle f(\vec e_j),\vec e^{\, i}\rangle $$ where $\vec e^{\, i}$ is the $i$-th element of the reciprocal basis of $\mathscr B$ so that it seems to me that generaly $$ a_{i,j}=\langle f(\vec e_j),\vec e^{\, i}\rangle\neq\langle f(\vec e_i),\vec e^{\, j}\rangle=a_{j,i} $$ that implies that $A$ is not symmetric. So could someone help me, please?

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No: the matrix of a Hermitian with respect to a basis that is not orthonormal is not necessarily symmetric.

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