An "apparent" contradiction for eigenvalues signs of $A=\left( \begin{array}{cc} a & -a \\ a-1 & 1-a \\ \end{array} \right)$. The following matrix
$$A=\left(
  \begin{array}{cc}
    a & -a \\
    a-1 & 1-a \\
  \end{array}
\right)$$
has eigenvalues $\lambda=0,1$ $\forall a\in\mathbb R$. Therefore $\lambda\geq 0$. 
So I would expect that
$$\lambda=\frac{\vec x^T A \vec x}{\vec x^T \vec x}\geq 0, \ \ \forall \vec x=(x,y)\in\mathbb R^2.$$
Namely,
$$\vec x^T A \vec x=x^2 a-xy+y^2(1-a), \ \ \ (1)$$
must be $\geq 0$. In order to study sign of $(1)$, I used the quadratic form theory, i.e. I study the associated symmetric matrix of $(1)$:
$$\tilde A=\left(
  \begin{array}{cc}
    a & -\frac{1}{2} \\
    -\frac{1}{2} & 1-a \\
  \end{array}
\right)$$
but it isn't a positive semi-definite matrix!
Someone help me to explain this "apparent" contradiction, please?

Checking that all eigenvalues are non-negative does not imply that a matrix is positive semidefinite.  This test only works if the matrix is Hermitian (symmetric in the real case), which the original matrix is not.
For a concrete counterexample, take $a=0, x=10, y=1$, and your quadratic form is negative.
