Is a positive operator symmetric? A question just came up in my mind right now, is a positive operator symmetric? I'm trying to prove this using the properties of positive operators, but I'm stuck, I need a hand here.
Thanks
 A: This drives me nuts. Every square matrix $A$ can be written as the sum of a symmetric matrix and a skew symmetric matrix. 
So, $$ A = B + C; \; \; B^T = B; \; \; C^T = - C. $$
Recipe: $$ B = \frac{1}{2}(A + A^T); \; \; C =   \frac{1}{2}(A - A^T). $$
We need to know that the transpose of a number, a one by one matrix, is itself.
Now, given a column vector $v,$ we find
$$  v^T C v  = (v^T C v)^T = v^T C^T v = - v^T C v. $$
So $$  v^T C v  = 0. $$
Then, $$ v^T A v = v^T B v + v^T C v = v^T B v.   $$
So, the quadratic form constructed from a (square) real matrix is defined entirely by the symmetric part. Positivity is determined by the symmetric part. 
For reasons entirely beyond my understanding, people discuss positivity for matrices that may not be symmetric. There is no real need to do this.  
AHHHHAAAAA!!!!  Horn and Johnson, Matrix Analysis (1985). This is pretty much the standard reference for matrices; check wikipedia. Page 397:

Exercise.  Show that if $A \in M_n $ is a real matrix and if $x^T A x$ is positive for all nonzero $x \in \mathbb R^n,$ then $A$
  need not be symmetric, and hence it need not be positive definite.
  Hint: Consider a real skew-symmetric matrix $A$ and compute $(x^TAx)^T.$ What is $x^T A x$ in this case? What about $x^\ast A x$ for
  nonreal $x?$

Note that for any matrix $P,$ the symbol $P^\ast$ is the conjugate transpose of $P.$
A: This is not necessarily true, consider the matrix $\left(\begin{array}{c c} 2 & 2 \\ 0 & 2\end{array}\right)$. Then, $$(x,y)\left(\begin{array}{c c} 2 & 2 \\ 0 & 2\end{array}\right)\left(\begin{array}{c} x \\ y\end{array}\right)=2x^2+2xy+2y^2=(x+y)^2+x^2+y^2,$$ so the operator $Tx=Ax$ is positive definite. However, $A$ is not symmetric.
