The task is to be solved outside the context of linear operators (we haven't learned them, but I think it might be closely related to that unit which is still to come). $\DeclareMathOperator{\rank}{rank}$
Let $A\in M_{mn}(\Bbb F), A\ne0$ and $\varphi:\Bbb R\to \Bbb R$ defined by:$$\varphi(t)=\rank(t\cdot A)$$ Is $\phi$ a continuous mapping?
My work: $$\rank(A)\le m,n\space\land\space\rank(t\cdot A)=\rank(A)\space\forall t\in \Bbb R\setminus\{0\}$$ or $$A\text{~}D_{\rank (A)}\implies (t\cdot A)\text{~}D_{\rank(A)} \implies\;\underbrace{[A]=[t\cdot A]}_{\text{equivalence classes}}$$ where $D_{\rank(A)}$ is a canonical matrix and $\rank$ $$\implies\varphi\;\text{ is constant on} \langle-\infty,0\rangle\cup\langle0,+\infty\rangle$$ In whole: $$\underset{A\in M_{mn}(\Bbb F)}\bigcup\mkern-12mu\mathcal R_\varphi=\Bbb N,\enspace\forall t\in \Bbb R\setminus \{0\}$$ My conclusion: $\varphi$ is discontinuous on $\Bbb R$ because it isn't continuous in every $t\in \mathcal D_\varphi$, more precisely in $t=0$.
Is this correct?
From the fact that $\phi|_{\mathbb{R}\setminus \{0\}}$ is a nonzero constant (equal to $\operatorname{rank}(A)$) and that $\phi(0) = 0$ it follows directly that $\phi$ is not continuous at $0$ and that it is continuous at every other $t \in \mathbb{R}$.
The assertion $\mkern-13mu\underset{A\in M_{mn}(\mathbb F)}\bigcup\mkern-12mu\mathcal R_{\phi}=\mathbb N$ is irrelevant as we are examining the continuity of $\phi$ for a fixed matrix $A\in M_{mn}(\mathbb F)$. Moreover, it is wrong, it should be $$\mkern-13mu\underset{A\in M_{mn}(\mathbb F)}\bigcup\mkern-12mu\mathcal R_{\phi}=\mathbb \{0, 1, 2, \ldots, \min\{m,n\}\}.$$