Self adjoint and symmetric operator

Question

I am wondering whether for an operator defined on a real Hilbert space to be positive we need to show that it is self-adjoint at first. It seems to me that they are two different property and can be studied independently.

Answer 1

In a complex Hilbert space, every positive operator is selfadjoint. But in a real Hilbert space, one can find positive operators which are not selfadjoint; for instance, $$ T=\begin{bmatrix}1&-1\\ 0&1\end{bmatrix} $$ as an operator on $\mathbb R^2$ satisfies $\langle Tx,x\rangle\geq0$ for all $x$.

This ambiguity is common in the definition of real positive-definite matrices, where some people just require the positivity condition, while others also require the matrix to be symmetric (i.e., selfadjoint since all entries are real).

Answer 2

Well, self adjoint implies $\langle Ax,x\rangle$ is real for all $x$ in the Hilbert space, and that better be the case if $\langle Ax,x\rangle \geq 0$ for all $x$.

In fact, positive operator implies self adjoint. Indeed, since $\langle Ax,x\rangle =\langle x,Ax\rangle=\langle A^*x,x\rangle$ for every $x$, we have $\langle (A^*-A)x,x \rangle =0$ for every $x$, which implies $A=A^*$.