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.

Solution 1:

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.

Solution 2:

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