Given matrices $A$ and $B$, how can I find a scalar $s$ that makes $A + s B$ rank-$1$? Given $3 \times 3$ matrices $A$ and $B$, how can I find a scalar $s$ that makes the matrix $A + s B$ rank-$1$? Is there a method using singular value decomposition or eigenvalues?
Thanks!
 A: Try Gauss elimination and pick $s$ that makes the resulting echelon matrix being of rank $1$ (if possible).
A: Generalized eigenvalue problem
$C = A + sB$ has rank 1 if and only if its range is one-dimensional, so its nullspace must be two-dimensional. I.e., there must exist two linearly independent vectors $x_1, x_2$ such that
$$0 = Cx_k = (A + sB) x_k \Leftrightarrow Ax_k = -sBx_k$$
for $k = 1,2$.
This is a generalized eigenvalue problem.
Rethinking a bit (or reading the link above), we see that for such an $s$ to exist, we must first have $\det (A + sB) = 0$, which might have several solutions. For a specific $s$ satisfying the determinant equation above, we can then calculate the nullspace of $A + sB$ and see if it has dimension two. This is very much like calculating regular eigenvalues and eigenvectors, just that you use the matrix $B$ instead of the identity matrix.
Minor determinants
This is another possible solution route.
The matrix $C = A = sB$ has rank one if and only if all determinants of $2 \times 2$ minors of the matrix is zero (the set of matrices with rank $\leq 1$ forms a variety, being the zero locus of the $2 \times 2$ minor determinants), and at least one entry in $C$ is non-zero. This gives you nine (second degree polynomial) equations in one unknown, $s$, that needs to be satisfied.
E.g. the equation for the minor of $C$ corresponding to row 1 and 2 and column a 1 and 2 will be:
$$\begin{align}
0 = &\begin{vmatrix}
a_{1,1} + s b_{1,1} & a_{1,2} + s b_{1,2} \\
a_{2,1} + s b_{2,1} & a_{2,2} + s b_{2,2}
\end{vmatrix} \\
&= (a_{1,1} + s b_{1,1})(a_{2,2} + s b_{2,2}) - (a_{2,1} + s b_{2,1})(a_{1,2} + s b_{1,2})
\end{align}$$
