\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?


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, $mn$.

  • $\begingroup$ so what I wrote is right, or did I get the columns $mn$ shape wrong $\endgroup$ – develarist Dec 4 '20 at 9:36
  • 1
    $\begingroup$ the column dimension is indeed $mn$, please see my edit $\endgroup$ – Christoph Hanck Dec 4 '20 at 10:39
  • $\begingroup$ in the answer to the following question, i try to correct someone else's take on $A'$ and $A''$'s column dimension. hope you also agree there, although I see now I misled them by incorrectly stating what $A$'s dimensions were in the comments due to a faulty source, which I then re-state in a later comment math.stackexchange.com/questions/3929230/… $\endgroup$ – develarist Dec 4 '20 at 10:43
  • $\begingroup$ Assuming you refer to the penultimate comment there, yes, the column dimension should also be $mn$. $\endgroup$ – Christoph Hanck Dec 4 '20 at 11:26

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