Normalization of conditional probabilities Let $p_i$ be a vector, such that $p_i\geq 0$ and $\sum_i p_i=1$. Similarly $\pi_\alpha$ is a vector satisfying $\pi_\alpha\geq 0$ and $\sum_\alpha \pi_\alpha = 1$. The two are related via $\pi_\alpha = \sum_i W_{\alpha i}p_i$, where $W_{\alpha i}\geq 0$. Can we make a mathematically rigorous statement that $\sum_\alpha W_{\alpha i}=1$?
Obviously, this is always the case when $p_i,\pi_\alpha$ are understood as probabilities and $W_{\alpha i}$ as conditional probabilities, in which case $W_{\alpha i}\geq 0$ is just the normalization of the conditional probability. Thus the question can be restated as: do we deal here with a simple mathematical fact or is there an additional implicit condition inherent in interpreting $W_{\alpha i}$ as probabilities.
Remark:  This might be a trivial question due to some glitch in my head, so a clarifying comment could suffice.
 A: Not necessarily. Here is a counter-example: consider the vectors
$$
p =   \begin{bmatrix}
   0.5  \\
   0.5   
 \end{bmatrix}
$$
and
$$
\pi =   \begin{bmatrix}
   0.5  \\
   0.5   
 \end{bmatrix}
$$
Then the matrix
$$
W =   \begin{bmatrix}
   1 & 0  \\
   1 & 0   
 \end{bmatrix}
$$
verifies $Wp = \pi$, while the elements in each of its columns do not sum to 1.
A: The answer by @CamilleGontier has pushed me in the right direction: what is implicit in the question (but was not explicitly stated in the OP) is that it should work for an arbitrary vector $p_i$ (as long as it satisfies conditions $p_i>0$ and $\sum_i p_i=1$.) We can then consider a set of linearly independent vectors which have only one non-zero component:
$$
p_i^{(i_0)}=\delta_{i,i_0}$$
Then for each such vector we have corresponding
$$
\pi_\alpha^{(i_0)}=W_{\alpha i_0},$$
and the normalization condition for $\pi_\alpha$ automatically translates into the condition for matrix $W$ :
$$\sum_\alpha\pi_\alpha^{(i_0)} = \sum_\alpha W_{\alpha i_0}=1
$$
