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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?

glS
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Fgilan
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  • If $V$ is a real vector space, then $\langle Tv,v \rangle$ can't be complex – Ben Grossmann Feb 15 '16 at 16:37
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    I notice you have asked 11 questions but you haven't accepted any answers. Do you know about this option? If not see here, and for a detailed discussion, here. Note, you don't have to accept an answer if you feel your question hasn't been fully dealt with, but in that case, you might want to comment on the existing answers, letting them know what you still don't understand. – Michael Albanese Feb 15 '16 at 22:44

3 Answers3

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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).

Sheldon Axler
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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).

Community
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Silvia Ghinassi
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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.

Roland
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    In general, writing $\langle Tv,v \rangle \geq 0$ holds automatically implies that the statement makes sense. – Silvia Ghinassi Feb 15 '16 at 16:39
  • @SilviaGhinassi: So you always consider ill-defined terms to be false? – Roland Feb 15 '16 at 16:40
  • I am not a logician, so I am entering a field I know nothing about, but in general I don't consider ill-defined terms in the first place, and I am under the impression textbooks tend to do the same. – Silvia Ghinassi Feb 15 '16 at 16:50