Show that a positive operator is also hermitian I'm having a little difficulty with this. Given some positive operator $A$, show that it is also hermitian.
(A positive operator is defined as $\langle Ax,x\rangle\ge 0$ for all $x \in V$ where $V$ is some vector space.)
Here's what I have so far.
We can construct $A = B + iC$ where $B,C$ are hermitian operators
$B = (A + A^*)/2$, $C = (-iA + iA^*)/2$ where $^*$ is the conjugate transpose.
I'm trying to show that $B$ and $C$ are diagonalizable by the same vectors, and that the eigenvalues of $C$ are $0$. I'm not sure how to do this though.
 A: Let's go this way. 
You already know how to show that any operator $A$ can be written as $A = B + iC$, where $B$ and $C$ are both Hermitian.
As $A$ is positive, for any $|v\rangle$ we should have $\langle v|A|v\rangle$ is a non-negative real number. As $B, C$ are Hermitian they have all real eigenvalues, and a spectral decomposition can be done. $B = \sum_j \lambda_j |j\rangle\langle j|$,$C = \sum_k \lambda_k |k\rangle\langle k|$, where $\lambda_j, \lambda_k \in \mathbb{R}$. Thus $\langle v|B + iC|v\rangle = \sum_j \lambda_j \langle v|j\rangle\langle j|v\rangle + \sum_k i\lambda_k\langle v|k\rangle\langle k|v\rangle$. 
As $\langle v|j\rangle\langle j|v\rangle$ is always non-negative real, the second part must always be 0. Thus $A = B$ is Hermitian
A: The following result is what you are trying to prove: 

If $V$ is a finite-dimensional inner product space over $\mathbb{C}$, and if $A: V \rightarrow V$ satisfies $\langle Av, v \rangle \geq 0$ for all $v \in V$, then $A$ is Hermitian.

The result is not true if $V$ is taken to be a real inner product space. That was the key missing ingredient from your question. Here are some strong hints to obtain the proof:


*

*Prove that, for all $v \in V$, $\langle (A - A^{\ast})v, v \rangle = 0$, by using the positivity assumption. Remember that over a complex space the inner product is conjugate-linear.

*Notice that $A - A^{\ast}$ is a normal operator. Then, by applying the spectral theorem, show that $A - A^{\ast}$ must in fact be the zero operator. 

A: A simpler way is as follows:


*

*Prove that if any operator $B$ satisfies $\langle x,By\rangle = 0$ for every $x$ and $y$ in the inner product space in $\mathbb{C}$, then $B=0$.

*Expand $\langle x+y,B(x+y)\rangle$ and $\langle x+iy,B(x+iy)\rangle$.


These two facts will show that if $B$ satisfies $\langle x,Bx\rangle = 0$ for every $x$ and it is a complex inner product space, then $B=0$. In our case, we take $B=A-A^*$ and conclude the result. Note that all operators are from $V\to V$. 
A: Since $A=B+iC$ with $B$ and $C$ hermitian, we can decompose the inner product thanks to the linearity in its second argument
$$
\langle \psi | A \psi \rangle=\langle \psi | (B \psi+iC \psi) \rangle=\langle \psi | B \psi \rangle+\langle \psi | iC \psi \rangle
$$
We can use the spectral decomposition: $B=Q^\dagger DQ$ and $C=Q'^\dagger D'Q'$ where $Q,Q'$ are orthogonal and $D,D'$ are diagonal matrices of real values.
$$
\langle \psi | B \psi \rangle+\langle \psi | iC \psi \rangle=
\langle \psi | Q^\dagger DQ \psi \rangle+\langle \psi | iQ'^\dagger D'Q' \psi \rangle
$$
Since the inner product is invariant with respect multiplication of orthogonal matrices
$$
\langle \psi | Q^\dagger DQ \psi \rangle+\langle \psi | iQ'^\dagger D'Q' \psi \rangle=
\langle \psi | D \psi \rangle+\langle \psi | iD' \psi \rangle
$$
If the $j$-th component of $|\psi\rangle$ is $(x_j+iy_j)$, the innero product $\langle \psi | D \psi \rangle$ can be written as
$$
b=\sum_j (x_j+iy_j)(x_j-iy_j)d_j=\sum_j (x_j^2+iy_j^2)d_j
$$
that is a real number.
Reasoning in a analogues way we have that $\langle \psi | iC \psi \rangle$ is a complex number, therefore it should be null.
We can conlcude that $A=B=Q^\dagger D Q$, i.e. $A$ is an hermitian operator.
