If $A, B$ are both $n×n$ column-stochastic matrices, then for any $w∈[0, 1], \ (1 − w)A + wB$ is also column-stochastic? 
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"? 
 A: 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.
A: 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.
