How to multiply a vector and a square matrix with Kronecker product, and know the answer's shape?

Question

• $\mathbf{1}_n \in \mathbb{I}^{n\times 1}$ is a vector of ones with shape $n\times 1$
• $\mathbf{I}_m \in \mathbb{I}^{m\times m}$ is an identity matrix with shape $m\times m$

What is the answer to, and shape of, $\mathbf{1}_n^\top \otimes \mathbf{I}_m$ ?

$$\mathbf{1}_n^\top \otimes \mathbf{I}_m = \overset{(1\times n)}{\begin{pmatrix} 1 & 1 \dots 1 \end{pmatrix}} \otimes \overset{(m\times m)}{\begin{pmatrix} 1 & 0 & \dots & 0 \\ 0 & 1 & \dots & 0 \\ \vdots &\vdots & \ddots & \vdots \\ 0 & 0 & \dots & 1 \end{pmatrix}} =? $$

Answer 1

It's $$\underbrace{[I_m|I_m|\cdots|I_m]}_{n \ \text{times}}$$

because is $A$ is $p \times q$ and $B$ is $m \times m$:

$$A\otimes B=\begin{pmatrix}a_{11}B&\cdots&a_{1q}B\\ \vdots&&\vdots\\ a_{p1}B&\cdots&a_{pq}B\end{pmatrix}\tag{1}$$

Therefore the shape of $\mathbf{1}_n^\top \otimes \mathbf{I}_m$ is $m \times mn$.