I'm interested in the following question:
Given that some matrix $M$ (which we have) can be expressed as $\vec{\lambda}\vec{x}^t + \vec{x}\vec{\mu}^t$, with $\vec{x}$ given, how can we determine values for $\vec{\lambda}$ and $\vec{\mu}$?
I'm interested in the general case when $\vec{x}$ is not necessarily perpendicular to $\vec{\lambda}$ or $\vec{\mu}$ (where this problem becomes finding the Singular Value Decomposition for this rank $2$ matrix).
Is there a systematic procedure for solving this problem?
Best Answer
Let me do it for a more general equation $\lambda x^t + y\mu^t = M$, with $x,y$ known.
We also assume that both $x$ and $y$ are non-zero.
Write $v = (1, 0, \dotsc, 0)^t$.
Since $x$ and $y$ are non-zero, we may then find invertible matrices $P, Q$ such that $Px = Qy = v$.
Multiplying the equation on the left by $Q$ and on the right by $P^t$, we get: $$(Q\lambda) v^t + v(P\mu)^t = QMP^t.$$
Writing $\lambda' = Q\lambda,\mu' = P\mu,M' = QMP^t$, we see that it suffices to solve the equation $\lambda' v^t + v \mu'^t = M'$.
But the vector $v$ is so simple that this equation is trivially solved: $\lambda'$ is just the first column of $M'$ (denoted by $c$), and $\mu'^t$ is just the first row of $M'$ (denoted by $r^t$), with the exception that the first components of $\lambda'$ and $\mu'$ are not uniquely determined. Instead, their sum is determined as the $(1, 1)$-th entry of $M'$ (denoted by $m$).
Hence we get a family of solutions $\lambda' = c - av$ and $\mu'^t = r^t - bv^t$, with $a + b = m$.
It only remains to multiply by the inverses of $P$ and $Q$ to get back $\lambda$ and $\mu$.