Real positive definite not self-adjoint operator

Question

I need find an example for a real bounded operator in a Hilbert space such that it is positive definite but not self-adjoint.
I think the left-shift operator in $l^2$ is not self-adjoint and positive-definite:
$$
L(x_1,x_2,\dots)=(x_2,x_3,\dots)\in l^2(\mathbf{R})
$$
taking two sequences :
$(0,1,0,\dots)$ and $(0,0,1,0,\dots)\in l^2$ we have
$$
\left<(0,1,0,\dots),L(0,0,1,0,\dots) \right>=1,
$$
however
$$
\left<L(0,1,0,\dots),(0,0,1,0,\dots)\right>=0.
$$
however I am not seeing the $\left<x,Lx\right> > 0$ condition for any non zero vector $x$.

Answer 1

The shift is not positive; for instance you have $$ \langle L(1,-1,0,\ldots),(1,-1,0,\ldots)\rangle=-1. $$ An easy example of a non-selfadjoint positive-definite operator on $\mathbb C^2$ is $$ \begin{bmatrix} 1&1\\0&1\end{bmatrix}. $$ If you want this on an infinite-dimensional Hilbert space, fix an orthonormal basis $\{e_n\}$ and put $$ Te_1=e_1,\qquad Te_2=e_1+e_2,\qquad Te_{k+2}=0,\ k\in\mathbb N. $$ Then for nonzero $x=\sum_kx_ke_k$ you have \begin{align} \langle Tx,x\rangle&=\langle (x_1+x_2)e_1+x_2e_2,x_1e_1+x_2e_2\rangle =|x_1|^2+|x_2|^2+x_2\overline{x_1}\\[0.2cm] &\geq \frac{|x_1|^2+|x_2|^2}2+\frac{|x_1|^2+|x_2|^2-2|x_2\overline{x_1}|}2\\[0.2cm] &= \frac{|x_1|^2+|x_2|^2}2+\frac{(|x_1|^2-|x_2|)^2}2\\[0.2cm] &\geq \frac{|x_1|^2+|x_2|^2}2>0. \end{align}