2
$\begingroup$

\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?

$\endgroup$
2
$\begingroup$

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, $mn$.

$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.