Can I Find the Eigenvalues of a Matrix this Way? I have a matrix $A = xx^T - yy^T,$ where both $x$ and $y$ are linearly independent $n$-column vectors, $n\geq 2$. To find the eigenvalues, I reasoned this way:
Since $x$ and $y$ are linearly independent vectors (given), then $rank(A) = 2.$ So, we have two non-zero eigenvalues and $(n-2)$ eigenvalues, each with a value of zero.
Since both $x$ and $y$ are linearly independent, they form a basis for $V = span(x,y).$ Therefore,
$Ax = (x.x)x - (x.y)y$ and
$Ay = (x.y)y - (y.y)y$
The matrix $A$ relative to $V$ is:
$A = \begin{bmatrix}x.x&x.y\\-x.y&-y.y\end{bmatrix}.$
Now,
$Av =$ $\lambda$$v$
Therefore, $(A - \lambda I_{2})v = 0$, where $A$ has been restricted to a $2 \times 2$ matrix.
When we solve this, we get the $\lambda$'s.
Is this correct so far?
Thanks.
 A: Yes, your solution is correct. The restriction of $A$ to $V = \operatorname{span}(x,y)$ is valid because $V$ is an $A$-invariant subspace of $\Bbb R^n$.
Here's an alternative solution that I like. The matrix $A$ can be expressed as the product $A = PQ$ where
$$
P = \pmatrix{x & y}, \quad Q = \pmatrix{x^T \\ -y^T}.
$$
Because $PQ$ and $QP$ have the same non-zero eiegenvalues, it suffices to find the eigenvalues of the matrix
$$
QP = \pmatrix{x^Tx & x^Ty\\ -y^Tx & -y^Ty},
$$
which confirms the validity of your approach.
A: You're right, this is an excellent solution. But probably we should add a little more. Like this.
Since $x,y$ are linearly independent, there exists a basis $V$ of the form $x,y,v_3,\ldots,v_n$. In this basis the matrix of the linear operator $z\to Az,\ z\in V$ has the form
$$
B=
\left(
  \begin{array}{rrrrr}
     x.x &  x.y & x.v_3  & \ldots & x.v_n \\
    -y.x & -y.y & -y.v_3 & \ldots & -y.v_n \\
       0 &    0 &      0 & \ldots &  0\\
  \ldots &\ldots&\ldots  &\ldots  &\ldots\\
       0 &    0 & 0      &\ldots & 0 \\       
  \end{array}
\right)
$$
and matrices $A$ and $B$ are similar.
