A "quadratic form" $x\mapsto x^TAx$ in which $A$ is not symmetric

Question

Let $n \in \{1,2,\dots\}$, and let $A$ be an $n\times n$ matrix of real numbers. Consider the unique function $f:\mathbb{R}^n\rightarrow\mathbb{R}$ satisfying $f(x) = x^TAx$ for every $x \in \mathbb{R}^n$. Suppose $A$ is not symmetric.

1. Is it possible for $f$ to be convex?
2. Suppose that $f(x) \geq 0$ for every $x \in \mathbb{R}^n$. Must $f$ be convex? What if $f(x) > 0$ for every $x \in \mathbb{R}^n$?
3. Is $A$ uniquely determined by $f$? In other words, suppose that $B$ is an $n\times n$ matrix of real numbers satisfying that $f(x) = x^TBx$ for every $x \in \mathbb{R}^n$; is it necessarily the case that $A=B$?

Answer 1

$$ x^TAx = \sum_{i,j}x_i A_{ij} x_j = \frac{1}{2}\sum_{i,j}x_i (A_{ij} +A_{ji})x_j = x^TBx, $$ where $$ B = \frac{A+A^T}{2} $$ is symmetric. Thus, all the properties true for a symmetrix $A$ matrix apply.