What does $\sum x$ mean when $x$ is a vector?

Question

I am reading a paper. They define the following problem. problem

Notice the equation $\sum p \cdot M = 1$. Now see the definition of $M$ in this snippet from the same paper: M

So, $M$ is some matrix with the same rows as the length of the $p$-vector, which we call $n$.

Then clearly $p\cdot M$ is itself a row vector of length $m$ (the column length of $M$). But then what is $\sum p \cdot M$? What is the sum of a vector? Have the authors made a mistake, or is this some notation I am unfamiliar with?

EDIT:

The answers say that this is just element-wise addition, but I do not see how this fits.

I will provide context. Consider this tree: tree Each arc in the tree is associated with a probability. Each node in the tree is the "starting point" of a distribution, corresponding to the arcs that branch off from this node. The first node has a distribution with two outcomes, and the other nodes have a distribution with 3 outcomes each.

$p$ is a vector with $8$ elements. It contains the probabilities on the arcs. $M$ is a matrix of $8$ rows and 3 columns (since we have 3 nodes/3 distributions). The idea is that $p \cdot M$, where $M$ is a matrix of zeros and ones, extracts probabilities for each distribution/for each column/for each node. In this context, how then do we make sense of $\sum pM = 1$?

How would you define $M$ in this case then?

Answer 1

The sum of a vector (at least in this case) is the sum of its entries.

Answer 2

From the little context given, one can imagine that in $\sum p\cdot M=1$, the $1$ in the RHS denotes a vector of ones.

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Update:

The explanation is given just above: "constraints defining the probabilities to be nonnegative and to sum up to one". The sum of the components of the vector $pM$ equals $1$.

Answer 3

Since you write that each column defines a distribution, then this is "somewhat equivalent" that each column is an observation / random variable.

Then, just like the standard axiom for probabilities:

$$\sum_i p_i x_i=1$$

One considers

$$\sum_{i,j} p_i x_{ij}=(1_1,1_2,...,1_{no. \space of \space cols})$$ where $j$ refers to the column and $i$ refers to the row. So for each column one sums over the rows and each such sum must evaluate to $1$.