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?
Best Answer
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).