Is self-adjoint operator in a Hilbert space is always positive operator? Let me first define the self-adjoint operator.

Let $A$ be a bounded operator in a Hilbert space $H$, then $A$ is said to be a self-adjoint operator if $A^*=A$.
And $A$ is known as a positive operator if $\langle Ax,x \rangle \geq 0$.

What I know is that a positive operator is not always a self-adjoint operator for example we can take the rotation operator in $\mathbb{R}^2$.
My question is whether the self-adjoint operator is always a positive operator?
According to me it is not because if $A$ is a self-adjoint operator then $\langle Ax,x \rangle$ is real and when this is greater than or equal to zero then we say it is a positive operator but I am finding it difficult to construct an example.
 A: Positive but not self-adjoint:
$T:\Bbb{R}^2\to \Bbb{R}^2$ defined by $T(x, y) =(x+2y, y) $
Then $T^{\star}(x, y) =(x, 2x+y)$
Then clearly $T\neq T^{\star}$ but $T$ is positive definite as
$$\begin{align}\langle T(x, y), (x, y) \rangle&=\langle (x+2y,y), (x, y) \rangle\\&=x^2+2xy+y^2\\&=(x+y) ^2\ge 0\end{align}$$

Self-adjoint but not positive:
Consider $T:\Bbb{R}^2\to \Bbb{R}^2$ defined by
$$T(x, y) =(x, -y) $$
Then $T^{\star}=T$ but
$\langle T(x, y) , (x, y) \rangle =x^2-y^2$
Hence $T$ is not positive semidefinite.

If $\mathcal{H}$ is complex Hilbert space then the answer is yes.

Theorem : Let $T\in \mathcal{L}({V_{\Bbb{C}}}) $ then $\langle Tv, v\rangle \in \Bbb{R}$ iff $T$ is self adjoint.
Then for positive operator $T$ as $\langle Tv, v\rangle\ge 0 $ implies $T$ is self adjoint.

A: Consider the Hilbert space $(\mathbb{R}^2,\cdot)$ with the dot product as the standard inner product. Then take $A = -I$, meaning $A$ is negative of the identity operator. It is clear that $A$ is bounded (try to prove this yourself). Moreover, $A$ is self adjoint since the matric of $A$ is real and symmetric, in particular, the matrix of $A$ equals its conjugate transpose. Yet, for the vector $v = (1,1)$ we have that $$Av \cdot v = (-1,-1) \cdot (1,1) = -2.$$ So, $A$ is not positive definite.
