Confusion in dimension of the vectors, and what is this $yA$ which is not $y^TA$

Question

This is from a paper by Laszlo Lovasz, can be found here, chapter 4. We have a $(0,1)$-matrix $$A = [a_{ij}]_{i,j=1}^{i=r,j=k}.$$ We then have 2 linear programs:

\begin{align}\min &\quad y\cdot 1\\ \text{s.t.}&\quad yA \geq w\\ &\quad y \geq 0,\end{align}

and

\begin{align}\max&\quad w\cdot x\\ \text{s.t.}&\quad Ax \leq 1\\ &\quad x \geq 0.\end{align}

Later it is shown that $x = (x_1,\ldots,x_k), w = (w_1,\ldots,w_k)$, which is natural. The bizarre thing is $y = (y_1,\ldots,y_k)$. What is this? I initially thought $yA$ is just a shorthand for $y^TA$, but now I am not sure. I thought that the first LP is the dual of the second LP, but, what?

Answer 1

There's a typo in the first occurrence of $y$. It's actually $y=(y_1,\dots,y_r)$ not $y_k$ and this is also used in all further occurrences of $y$ in that paper.

So your initial guess of $yA$ being short for $y^TA$ appears to be correct.

I guess they just omit the transpose sign and always assume matching dimensions. Only the order of the factors in the matrix-vector-product reveals what kind of vector that is.