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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.

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    $\begingroup$ $A= -\mathrm{Id}$ is a counter-example $\endgroup$
    – Didier
    Jul 18, 2022 at 18:21

2 Answers 2

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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.


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    $\begingroup$ Aren't you answering a different question? $\endgroup$ Jul 18, 2022 at 18:06
  • $\begingroup$ Thanks, But actually the question was "Is a self-adjoint operator a positive operator or not?" $\endgroup$ Jul 18, 2022 at 18:09
  • $\begingroup$ @paulgarrett Thanks. I haven't read the question carefully. But now the problem is fixed. $\endgroup$ Jul 18, 2022 at 18:20
  • $\begingroup$ @AmitVishwakarma I have edited my answer. Now it looks ok. Isn't it? $\endgroup$ Jul 18, 2022 at 18:22
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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.

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