I do not understand why the equation below holds assuming $A$ and $P$ are both square matrices in $\Bbb{R}^{n*n}$ and $P$ is symmetric and positive definite (i.e. $ P = P^{T} $ and $ x^{T}Px > 0 $)
$$ 2 PA = A^{T}P + PA $$
I am trying to understand the LQR from the following slides (sheet 2). I can understand why $ \partial V(x)f(x) = 2x^{T}PAx $, but I do not understand why $ 2x^{T}PAx = x^{T}[A^{T}P+PA]x $. I think it has something to do with $ 2PA = A^{T}P + PA $, but I am not sure.
On
$x^TAx=x^TBx\quad \forall x\in\mathbb{R}^n\quad$ does not imply $A=B$ but it implies that the symmetric part of $A$ equals the symmetric part of $B$.
The proof of this is very simple of this is very simple by splitting the matrix into symmetric and anti symmetric parts and it is well known that the quadratic equation of an anti-symmetric matrix is 0.
Because $x^T PA x$ is a scalar, its transpose is equal to itself, i.e. $x^T PA x = (x^T PA x)^T = x^T A^T P x$. But $PA \neq A^T P$ in general.