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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 $x$.

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  • $\begingroup$ Plane rotation of acute angle $\endgroup$ Commented Jun 3, 2023 at 22:38
  • $\begingroup$ Take any real positive definite symmetric matrix $A$ and modify two terms $a_{12}+1=:\tilde{a}_{12}$ and $a_{21}-1=:\tilde{a}_{21}.$ The new matrix $\tilde{A}$ is not symmetric but its quadratic form coincides with that of $A.$ $\endgroup$ Commented Jun 3, 2023 at 23:20
  • $\begingroup$ Notation comment: the way notation is commonly used, $\ell^2(\mathbb R)$ does not mean what you think it means. The $\mathbb R$ indicates the domain and not the codomain. The codomain is usually implied. So $\ell^2(\mathbb R)$ denotes the space of square summable functions $\mathbb R\to\mathbb R$. The set of real/complex square summable sequences is commonly denoted by $\ell^2(\mathbb N)$. $\endgroup$ Commented Jun 4, 2023 at 11:16

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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}

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  • $\begingroup$ I think it was possible because this is over real. If it were over complex, it might be impossible. $\endgroup$
    – Balaji sb
    Commented Jun 4, 2023 at 5:25
  • $\begingroup$ Note that nothing in my answer changes if I replace $\mathbb C$ with $\mathbb R$. $\endgroup$ Commented Jun 4, 2023 at 11:10
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A bounded operator $A$ on a real Hilbert space is positive definite if and only if $A+A^t$ is positive definite as $$\langle (A+A^t)x,x\rangle =2\langle Ax,x\rangle $$ In particular any skew symmetric operator ($A^t=-A$) is positive semidefinite.
The identity operator is obviously positive definite, as $\langle Ix,x\rangle =\|x\|^2.$ Hence $I+B$ is positive definite for any skew symmetric matrix $B,$ as $$(I+B)^t+I+B=2I$$

Basing on the above it is easy to come up with a nonsymmetric positive definite operator. For example consider $\mathbb{R}^2$ and $$A =I+\begin{pmatrix} 0 & -1\\ 1 &0\end{pmatrix}=\begin{pmatrix} 1& -1\\ 1& 1\end {pmatrix} $$ Then $A+A^t=2I.$ Observe that $2^{-1/2}A$ corresponds to the rotation by the angle ${\pi/4},$ so it fits into the class of examples proposed by @AnneBauval in the comment section. The example on $\mathbb{R}^2$ can be extended to $\ell^2$ by $$Ax=(x_1-x_2,x_1+x_2,x_3,x_4,\ldots)$$ Then $A+A^t=2I.$

Remark If $A$ is a positive definite operator and $\langle Av,v\rangle = 0$ then $(A+A^t)v=0.$ Indeed $$0\le \langle A(\alpha v+w),\alpha w+v\rangle =\alpha \langle (A+A^t)v,w\rangle +\langle Aw,w\rangle $$ for all real $\alpha $ and all $w.$ The last expression is a nonnegative affine function of the variable $\alpha,$ thus $\langle (A+A^t)v,w\rangle=0$ for all $w.$ Hence $ (A+A^t)v=0.$

For the shift operator $S$ we have $\langle Se_1,e_1\rangle =0$ and $(S+S^t)e_1=e_2.$ Thus $S$ is not positive definite.

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