The sign of eigenvalues of $xx^T-yy^T$ Let $x,y \in \mathbb{R}^n$, I noticed that $(xx^T-yy^T)$ can have at most two non-zero eigenvalues and they have different signs or it can have just one non-zero eigenvalue which is positive.
1- Is the reason for which $(xx^T-yy^T)$ is a rank-2 matrix the fact that it's a span of only two vectors?
2- Is the only possible way for $(xx^T-yy^T)$ to have one non-zero eigenvalue is having the vectors in the same direction? then why?
3- Why the sign of the non-zero eigenvalues are different?
 A: Let $A=xx^T-yy^T$. Note that $A$ is symmetric, so it is diagonalizable over $\mathbb R$.
1- Yes. More explicitly, for any $z\in\mathbb R^n$ we have
$$
  Az=(x\cdot z)x-(y\cdot z)y\in\mathrm{span}(x,y),
$$
so $\mathrm{im}\,A\subseteq\mathrm{span}(x,y)$ is at most two dimensional.
2- Suppose $x$ and $y$ are linearly independent, so they form a basis for $V=\mathrm{span}(x,y)$. Consider the effect of $A$ on this basis:
$$
  Ax=(x\cdot x)x-(x\cdot y)y,
$$
$$
  Ay=(x\cdot y)x-(y\cdot y)y.
$$
Thus the matrix of $A|_V$ relative to this basis is
$$
  \begin{bmatrix}x\cdot x&x\cdot y\\-x\cdot y&-y\cdot y\end{bmatrix}.
$$
The determinant is
$$
  -\|x\|^2\|y^2\|+(x\cdot y)^2=-\|x\|^2\|y\|^2\sin^2\theta
$$
where $\theta$ is the angle between $x$ and $y$. Since they are linearly independent, $\sin^2\theta>0$, so the above $2\times2$ matrix has full rank. Thus $A|_V$ has two nonzero (and distinct; see below) eigenvalues.
These are also eigenvalues of $A$, since the eigenvectors of $A|_V$ are just the eigenvectors of $A$ which happen to lie in $V$. This shows $A$ has at least two nonzero eigenvalues. It must then have exactly two since $A$ has rank at most $2$.
To see this another way, we can find a basis of $\mathbb R^n$ whose first two elements are $x$ and $y$. With respect to this basis, the matrix for $A$ has the form
$$
  \begin{bmatrix}
    x\cdot x&x\cdot y&*&\cdots&*\\
   -x\cdot y&-y\cdot y&*&\cdots&*\\
   0&0&0&\cdots&0\\
   \vdots&\vdots&\vdots&\ddots&\vdots\\
   0&0&0&\cdots&0
  \end{bmatrix}.
$$
This is a block upper triangular matrix, so the set of eigenvalues of $A$ is the union of the sets of eigenvalues of the blocks.
3- Continuing from the above, since the determinant of $A|_V$ is negative, the two eigenvalues must have opposite sign.
