# Real positive definite not self-adjoint operator

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 $$\left=0.$$ however I am not seeing the $$\left > 0$$ condition for any non zero vector $$x$$.

• Plane rotation of acute angle Jun 3 at 22:38
• 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 $A.$ Jun 3 at 23:20
• 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 $\ell^2(\mathbb N)$. Jun 4 at 11:16

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}

• I think it was possible because this is over real. If it were over complex, it might be impossible. Jun 4 at 5:25
• Note that nothing in my answer changes if I replace $\mathbb C$ with $\mathbb R$. Jun 4 at 11:10

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.