Diagonalization of $M=ab^t+ba^t$ Given $a=(a_i)_{i=1}^n$ and $b=(b_i)_{i=1}^n$ vectors in $\mathbb{R}^n$, we define the matrix $M=(m_{ij})_{i,j=1}^n$ as:
$$
m_{ij}=a_ib_j + a_jb_i,
$$
or equivalently 
$$
M=ab^t+ba^t.
$$ 
Note that $a,b$ are also $n\times 1$ matrices, $a^t,b^t$ are $1\times n$ matrices.
What are the eigenvalues and eigenvectors of $M$?
 A: We are considering the following two cases:
Case I. $a$, $b$ linearly dependent. Let for example $b=\mu a$, then $M=2\mu\, a a^t$. But the rank of $aa^t$ is 1, as all its columns are multiple of the first column (if the first is nonzero.), and hence its eigenvectors are $a$, with eigenvalue $\lambda= 2\mu\, a^ta$, and $n-1$ eigenvectors corresponding to $\lambda=0$, which constitute a basis of the kernel of $aa^t$.
Case II. If $a$, $b$ linearly independent (and hence ∥a∥,∥b∥≠0). Then 
$$
Mx=\lambda x \,\,\Longleftrightarrow\,\, a(b^tx)+b(a^tx)=\lambda x,\tag{1}
$$
thus, if $\lambda\ne0$, then $x$ is a linear combination of $a$ and $b$. Hence $M$ has as eigenvalues $\lambda=0$ with multiplicity $n-2$, and eigenspace
$$
\{x:a^tx=b^tx=0\}.
$$
Also, setting $x=c_1a+c_2b$ in $(1)$ we obtain the system
$$
a^ta c_1+a^tb c_2=\lambda c_1, \quad a^tb c_1+b^tb c_2=\lambda c_2,
$$ 
which provides another two eigenvalues/eigenvectors. The eigenvalues satisfy the equation
$$
0=\left|
\begin{matrix}a^ta-\lambda & a^tb \\ a^tb & b^tb-\lambda \end{matrix}
\right|=\lambda^2-(|a|^2+|b|^2)\lambda+(|a|^2|b|^2-(a^tb)^2),
$$
and thus
\begin{align}
\lambda_{1,2}&=\frac{|a|^2+|b|^2\pm\sqrt{(|a|^2+|b|^2)^2-4(|a|^2|b|^2-(a^tb)^2)}}{2} \\
&=\frac{|a|^2+|b|^2\pm\sqrt{(|a|^2-|b|^2)^2+4(a^tb)^2}}{2}.
\end{align}
A: Some small notes, since Yiorgos' has done a fairly thorough job:
The key insight here is that $M$ can be written very nicely as the sum of rank-1 matrices.
That is: if we take $a,b$ as vectors in $\mathbb{R}^n$, $ab^T$ gives us the matrix whose entries are $M'_{ij} = a_i b_j$, and $b a^T$ gives us the matrix whose entries are $M''_{ij} = a_j b_i$.  Noting that $M_j = M'_j + M''_j$, it follows that $M = ab^T + ba^T$, and Yiorgos' analysis follows.
Note also that for vectors $u,v \in \mathbb{R}^n$, we can express the dot-product as
$$
u \cdot v = v^T u = u^T v
$$
Hence, the set
$$
\{x:a^Tx=b^Tx=0\}.
$$
is the set of all $x \in \mathbb{R}^n$ that are orthogonal to both $a$ and $b$.

So, suppose that $a,b$ are linearly independent.  We note that we may select $n-2$ eigenvectors corresponding to $\lambda = 0$ by selecting an orthogonal basis for the subspace
$$
\{x:a^Tx = b^Tx = 0\} = [\operatorname{span}(a,b)]^T
$$
The remaining 2 eigenvalues and eigenvectors, we note that because $M$ is symmetric, we may select a set of mutually orthogonal eigenvectors, so that two mutually orthogonal eigenvectors can necessarily be selected from the set
$$
\operatorname{span}(a,b) = \{c_1 a + c_2 b : c_1,c_2 \in \mathbb{R} \}
$$
So, consider any $x$ that can be written as $x = c_1a + c_2 b$, and suppose that $x$ is an eigenvector for some $\lambda$.  We have
$$
a b^T (c_1 a + c_2 b) + b a^T(c_1 a + c_2 b) = \lambda(c_1 a + c_2 b)
$$
We may rearrange the equation to get
$$
[(b^T a - \lambda) c_1 + b^T b c_2] a + [a^T a c_1 + (a^T b - \lambda) c_2] b = 0
$$
Since $a$ and $b$ are linearly independent, this gives us a system of equations for $c_1$ and $c_2$:
$$
(b^T a - \lambda) c_1 + (b^T b) c_2 = 0\\
(a^T a) c_1 + (a^T b - \lambda) c_2 = 0
$$
Or in other words, the set of $(c_1,c_2)$ corresponding to eigenvectors of the original matrix are precisely the eigenvectors of the new matrix
$$
\pmatrix{b^T a & b^Tb\\ a^T a & a^T b}
$$
