Is the self-adjoint condition required in the definition of a positive operator?

Question

I'm reading Linear Algebra Done Right and it defines a positive operator $T$ as one which is self adjoint and has the property $$\langle Tv,v \rangle \geq 0$$ for all $v\in V$.

I am confused as to why the self adjoint condition must be included. Here is what I came up with:

Suppose $T$ is an operator such that $\langle Tv, v\rangle \geq 0$ for all $v$. This implies that $\langle Tv, v\rangle$ is a real number, since the greater than sign doesn't make sense for complex numbers. Then, using the definition of adjoint, $$\langle Tv, v\rangle = \langle v, T^*v\rangle = \overline{\langle T^*v,v\rangle} = \langle T^*v, v\rangle$$ for all $v\in V$. Therefore, $Tv=T^*v$ for all $v$ and $T$ is self adjoint.

Where did I go wrong?

Answer 1

Your proof is correct in the complex case, which seems to be the case you have. You are correct that you don't need to assume self-adjointness for a complex positive operator (in the real case, knowing $\langle Tx,x \rangle \in \mathbb R$ is not very useful) as it follows from the positiveness

Notice that the conclusion that $\langle Tv,v\rangle= \langle T^*v,v\rangle$ actually implies $T=T^*$ is non-trivial (you can take a look at this question: Proof Complex positive definite => self-adjoint).

Answer 2

Your argument is "If $\langle Tv,v \rangle$ wouldn't be real, we couldn't write down $\langle Tv,v \rangle \geq 0$."

The line of thought behind requiring $T$ to be self-adjoint is "Only if we make sure that $\langle Tv,v \rangle$ is real, we can write down $\langle Tv,v \rangle \geq 0$.".

Your point of view is using the inequality as an implicit definition, whereas the other way explicitely ensures that you don't write down a nonsensical equation.

Answer 3

As stated in Linear Algebra Done Right immediately after the definition of a positive operator, the requirement that $T$ is self-adjoint can be dropped from the definition in the case of a complex inner-product space. However, the self-adjoint condition is needed on real inner-product spaces. Consider, for example, the operator $T$ on $\mathbf{R}^2$ of rotation by $90^\circ$. For this operator $T$ we have $\langle Tv, v \rangle \ge 0$ for all $v \in \mathbf{R}^2$ (because $\langle Tv, v \rangle = 0$ for all $v \in \mathbf{R}^2$), but $T$ is not self-adjoint and $T$ definitely should not be considered to be a positive operator (it has no real eigenvalues).