If $A, B$ are both $n×n$ column-stochastic matrices, then for any $w∈[0, 1], \ (1 − w)A + wB$ is also column-stochastic?
The solution seems trivial; each column must sum to one, and $1 \cdot w$ + $1 \cdot (1-w)$ = $1$
I'm certain that I'm missing something obvious, though. What is wrong with this trivial "proof"?
Let
$A = [a_{ij}] \tag 1$
and
$B = [b_{ij}]; \tag 2$
the the hypothesis that $A$ and $B$ are column-stochastic says that
$\forall i, j \;\; 0 \le a_{ij} \le 1; \;\forall i, \; \displaystyle \sum_j a_{ij} = 1; \tag 3$
$\forall i, j \;\; 0 \le b_{ij} \le 1; \; \forall i, \; \displaystyle \sum_j b_{ij} = 1; \tag 4$
then for $w \in [0, 1]$, $1 - w \in [0, 1]$ as well, whence
$0 \le w a_{ij} + (1 - w)b_{ij} \le w + (1 - w) = 1, \tag 5$
since we have assumed in (2), (3) that $0 \le a_{ij}, b_{ij} \le 1$; this shows us that
$0 \le (wA + (1 - w)B)_{ij} \le 1; \tag 6$
also
$\displaystyle \sum_j (wA + (1 - w)B)_{ij} = \sum_j (w a_{ij} + (1 - w) b_{ij})$ $= \displaystyle w \sum_j a_{ij} + (1 - w) \sum_j b_{ij} = w + (1 - w) = 1; \tag 7$
conditions (6) and (7) in concert are precisely the requirements that $wA + (1 - w)B$ be column stochastic. And we are done.
It is in fact just about as simple as our OP Arbalester surmised.
Sometime we get lucky and the useful stuff comes easy.
The following excerise is easy to show:
If $A$ is an entry-wise nonnegative matrix, then $A$ is column stochastic if and only if $e^\top A = e^\top$.
Thus, if $A$ and $B$ are both entry-wise nonnegative matrices and $\alpha \in [0,1]$, then $\alpha A + (1-\alpha)B \ge 0$ and $$ e^\top \left( \alpha A + (1-\alpha)B \right) = \alpha e^\top A + (1-\alpha)e^\top B = \alpha e^\top+(1-\alpha)e^\top = e^\top, $$ and the result is established.
