# 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.

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

## 2 Answers

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.

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

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.