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?

Suppose $A$ has real, positive eigenvalues and real eigenvalues.  Then the other answers are correct, but here is some geometric intuition as to why $x^T A x<0$ is possible if and only if $A$ is non-Hermitian.  The sign of $x^T A x$ is negative iff the angle between $A x$ and $x$ is more than $90^\circ$.  Since $A$ has positive eigenvalues, $A x$ and $x$ are certainly in the same "quadrant", as defined by the eigenvectors.  For a Hermitian matrix $A$, the eigenvectors are orthogonal, so each quadrant spans $90^\circ$, and $A x$ and $x$ can't be more than $90^\circ$ apart.  But for a non-Hermitian matrix $A$, the eigenvectors need not be orthogonal, in which case two of the quadrants span more than $90^\circ$, and $A x$ and $x$ can be more than $90^\circ$ apart even while lying in the same "quadrant".
