# Is the matrix of an hermitian (real) endomorphism symmetric?

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?

## 1 Answer

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