Show that $A$ is self-adjoint if and only if $(Ax,x)$ is real for all x

Question

Let $A:X\rightarrow X$ be a linear transformation, where $X$ is a finite-dimensional complex inner product space. Show that $A$ is self-adjoint if and only if $(Ax,x)\in\mathbb{R}$ for all $x\in X$.

Answer 1

First let we suppose that $A$ is self-adjoint then, \begin{eqnarray*} (Ax,x)=(x,A^{*}x)=(x,Ax)=(\overline{Ax,x}) \end{eqnarray*} which proofs that $(Ax,x)$ is real for all $x\in X$. Now let us prove that is sufficient. Let $(Ax,x)\in\mathbb{R},~\forall x\in X$, \begin{eqnarray*} (Ax,x)&=&(x,A^{*}x)=(\overline{A^{*}x,x})=(A^{*}x,x),~\forall x\in X\\ 0&=&(C=(A-A^{*})x,x) \end{eqnarray*} then all we need to prove is that, $C\equiv 0$ first apply $C$ in $x+y$ to get
\begin{eqnarray} 0&=&(C(x+y),x+y)\Rightarrow (Cx,y)+(Cy,x)=0\\ 0&=&(C(ix+y),ix+y)\Rightarrow (Cx,y)-(Cy,x)=0 \end{eqnarray} then $$(Cy,x)=0,~\forall x,y\in X\Rightarrow C\equiv 0$$ or more precisely $A=A^{*}$.