Bound the rank of $u\otimes A\in\mathbb R\otimes\mathbb R^m\otimes\mathbb R^n$ by the rank of $A$

Question

Let $l,m,n\in\mathbb N$, $u\in\mathbb R$, $A\in\mathbb R^m\otimes\mathbb R^n$ and $B:=u\otimes A\in\mathbb R\otimes\mathbb R^m\otimes\mathbb R^n$. I want to show a hilariously simple thing: Assuming the rank of the matrix$^1$ $A\in\mathbb R^{m\times n}$ is at most a prescribed $r\in\mathbb N$, then the rank of the matrix $B\in\mathbb R^{(1\times m)\times n}$ is at most $r$ as well.

How can we do that?

It's hard for me to not getting confused by the involved reindexing. Can we express the range of $B$ in terms of the range of $A$?

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$^1$ Remember that if $H_i$ is a $\mathbb R$-Hilbert space with $d_i:=\dim H_i\in\mathbb N$, we have canonically isomorphisms $$H_1\otimes H_2\cong\mathfrak L(H_1,H_2)\cong\left(\mathbb R^{d_2}\right)^{d_1}\cong\mathbb R^{d_1\times d_2}\tag1.$$

Answer 1

We have for general linear operators $T_i \in \mathrm{Hom}(V_i,W_i)$ that $$\mathrm{rank} \left(\otimes_{i=1}^n T_i\right) = \prod_{i=1}^n \mathrm{rank}(T_i).$$

This can for example be found in the lecture notes by Tin-Yau Tam (theorem 3.4.2).

Thinking of $u$ as a $1 \times 1$ matrix yields your bound.