<s>[INST] How to find the eigenvalues of $x x^T - y y^T$? 
If $x$ and $y$ are two linearly independent column $n$-vectors, where $n \geq 2$, find all the eigenvalues of $x x^{T} - y y^{T}$


Could you please tell me whether what I have written is correct or not?
Using the identity
\begin{align*}
\lambda^n\det(\lambda I_{(m)} - AB) = \lambda^m\det(\lambda I_{(n)} - BA)
\end{align*}
for $A \in F^{m \times n}$ and $B \in F^{n \times m}$, we can calculate the characteristic polynomial of $xx^T - yy^T$ by setting $A = (x, y) \in F^{n \times 2}$ and $B = (x^T, -y^T)^T \in F^{2 \times n}$ directly as:
\begin{align*}
\varphi(\lambda) &= \det(\lambda I_{(n)} - (xx^T - yy^T)) = 
\lambda^{n - 2}\det\left(\lambda I_{(2)} - \begin{pmatrix} x^T \\ -y^T \end{pmatrix}\begin{pmatrix} x & y \end{pmatrix}\right) \\
&= 
\lambda^{n - 2}\begin{vmatrix}
\lambda - x^Tx & -x^Ty \\
y^Tx & \lambda + y^Ty
\end{vmatrix} \\
&= \lambda^{n - 2}[(\lambda - x^Tx)(\lambda + y^Ty) + (x^Ty)^2] \\
&= 
\lambda^{n - 2}(\lambda^2 - (x^Tx - y^Ty)\lambda - (x^Txy^Ty - (x^Ty)^2))
\end{align*}
Since $x$ and $y$ are linearly independent, by Cauchy-Schwarz inequality $(x^Tx)(y^Ty) > (x^Ty)^2$ (that is, the equality of C-S inequality cannot hold), whence the determinant $\Delta$ of the quadratic equation $\lambda^2 - (x^Tx - y^Ty)\lambda - (x^Txy^Ty - (x^Ty)^2) = 0$ equals to
\begin{align*}
\Delta = (x^Tx - y^Ty)^2 + 4(\|x\|^2\|y\|^2 - (x^Ty)^2) > 0.
\end{align*}
Hence the two non-zero eigenvalues are two distinct real numbers
\begin{align*}
\lambda_1 = \frac{y^Ty - x^Tx + \sqrt{\Delta}}{2}, \quad 
\lambda_2 = \frac{y^Ty - x^Tx - \sqrt{\Delta}}{2}.
\end{align*}
 [/INST] Every vector simultaneously orthogonal to $x$ and $y$ is an eigenvector of the matrix $xx^T - yy^T$ of eigenvalue 0 (note that the matrix has rank 2, so his kernel must have dimension $n-2$ and this is always an eigenspace). So, the remaining eigenvectors must be linear combination of $x$ and $y$.
Solve
$$ (xx^T - yy^T)(\alpha x + \beta y) = \lambda (\alpha x + \beta y) \qquad \lambda,\alpha,\beta \in \mathbb{R}. $$
You will find two polinomial equation in $\lambda,\alpha,\beta$. It is easy to see that it can't be $\alpha=0$ and $\beta=0$. So, from the two equation you can find a polinomial equation of degree 2 in $\frac{\alpha}{\beta}$. Then you can compute the two eigenvalues and the corrisponding eigenvectors.
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