Eigenvalue and outer-products I am interested in the eigenvalues of:
$$aa^H+bb^H$$
where $a$ and $b$ are Nx1 dimensional complex vectors, assumed not to be colinear. This resulting matrix will be rank 2 and Hermitian. I can diagonalize it as
$$aa^H+bb^H=
\begin{bmatrix}
    v_1 & v_2 \\
    \vdots & \vdots  
\end{bmatrix}
\begin{bmatrix}
    \lambda_1 & 0 \\
    0 & \lambda_2  
\end{bmatrix}
\begin{bmatrix}
    v_1^H & \ldots \\
    v_2^H & \ldots  
\end{bmatrix}
$$
which gives an NxN matrix where $v_1$ and $v_2$ are the two orthonormal eigenvectors (with nonzero eigenvalues -  there are N-2 zero eigenvalues). I know that these eigenvalues are same as the eigenvalues of
$$
\begin{bmatrix}
    a^H & \ldots \\
    b^H & \ldots  
\end{bmatrix}
\begin{bmatrix}
    a & b \\
    \vdots & \vdots  
\end{bmatrix}
=
\begin{bmatrix}
    a^Ha & a^Hb \\
    b^Ha & b^Hb  
\end{bmatrix}
$$
But this is a 2x2 matrix. Why do the eigenvalues of these two matrices match (ignoring nonzero eigenvalues)?
What are some keywords I can read up.
 A: Let $M = aa^H + bb^H$. Note that, for any $x$,
$$Mx = aa^Hx + bb^Hx = (a^Hx)a + (b^Hx)b \in \operatorname{span}\{a, b\},$$
so the eigenvectors for non-zero eigenvalues must lie in this span.
Suppose that $\alpha a + \beta b$ is an eigenvector for $M$, corresponding to eigenvalue $\lambda$. Then,
$$(a^H(\alpha a + \beta b))a + (b^H(\alpha a + \beta b))b = \lambda (\alpha a + \beta b).$$
If $a$ and $b$ are linearly independent, as assumed, then uniqueness of linear combinations implies
\begin{cases}
(a^Ha)\alpha + (a^Hb) \beta &= \lambda \alpha \\
(a^Hb)\alpha + (b^Hb) \beta &= \lambda \beta.
\end{cases}
Note: this corresponds to finding the eigenvalues of the $2 \times 2$ matrix you specified.
(There's probably a more enlightening way to show this, but this is clean and straightforward.)
A: $
\def\l{\lambda}
\def\c#1{\color{red}{#1}}
$Let $A\in{\mathbb R}^{m\times n}$ be a rectangular matrix $(m>n)$, then the eigenvalue equation for the matrix $A^H\!A$ is
$$\eqalign{
A^H\!A v &= \l v
}$$
Multiplying by $A$ yields
$$\eqalign{
AA^H\,\c{Av} &= \l\,\c{Av} \\
AA^H\,\c{w} &= \l\,\c{w} \\
}$$
which is the eigenvalue equation for $AA^H$ with those same eigenvalues.
$\big($Keep in mind that $AA^H$ will have additional zero eigenvalues$\big)$
In your particular case, $\,n=2\;\,$ and $\;A={\tt[}\,a\;\;b\,{\tt]}$
