Resulting shapes when partitioning the constraint matrix $\boldsymbol{A}$ in linear programming

Question

\begin{equation} \boldsymbol{A} = \begin{bmatrix} {1}_n^\top \otimes \mathbb{I}_m \\ \mathbb{I}_n \otimes {1}_m^\top \end{bmatrix} \in \mathbb{R}^{(m+n)\times mn} \end{equation}

If the above matrix is partitioned as follows, are the dimensions shown below correct?

\begin{equation} \boldsymbol{A}' = \begin{bmatrix} {1}_n^\top \otimes \mathbb{I}_m \end{bmatrix} \in \mathbb{R}^{m\times mn} \end{equation}

\begin{equation} \boldsymbol{A}'' = \begin{bmatrix} \mathbb{I}_n \otimes {1}_m^\top \end{bmatrix} \in \mathbb{R}^{n\times mn} \end{equation}

If not, have I misinterpreted the steps to partitioning a block matrix? Can you then show how it can be corrected?

Answer 1

I would assume that $\otimes$ is the Kronecker product and that $A'$ refers to the first set of rows of $A$ (and not a transpose, as that notation is also often used - so that is not your question, I assume), i.e., that

$$ A=\begin{pmatrix}A'\\A''\end{pmatrix} $$ Also, $1_n$ is an $n$-dimensional column vector, so that $1_n^\top$ is $(1\times n)$.

Then, the results follow from the definition of a Kronecker product, which say that the row dimension of $C\otimes D$ is the product of the row dimensions of $C$ and $D$, and likewise for the column dimension. Since $1_n^\top$ has row dimension 1 and $\mathbb{I}_m$ has row-dimension $m$, ${1}_n^\top \otimes \mathbb{I}_m$ has row-dimension $m$. The same reasoning leads to $n$ rows for $A''$, so $n+m$ rows in total.

Since $1_n^\top$ has $n$ columns and $\mathbb{I}_m$ has $m$, the column dimension is, analogously, $mn$.