Let $B\in\mathbb{R}^{n\times m}$ and $A=\lambda I$, for $\lambda\in\mathbb{R}$. I want to show that a necessarily condition for the controllability of $(A,B)$ is $m\ge n$.
I assume this means for $$A=\begin{bmatrix}\lambda_{1} & 0 & \cdots & 0 \\ 0 & \lambda_{2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_{n}\end{bmatrix} \qquad B=\begin{bmatrix} b_{11} & \cdots & b_{1m} \\ \vdots & \ddots & \vdots \\ b_{n1} & \cdots & b_{nm}\end{bmatrix}$$
Is it sufficient to state that the controllability matrix cannot be defined for $m<n$, since in order to compute the controllability matrix, we must compute $A\cdot b$, which cannot be done if $m<n$?...Or maybe I've misinterpreted the question?
I believe your interpretation of the question may be incomplete. It is usually assumed that $ A \in \mathbb{R}^{n \times n} $ and $ B \in \mathbb{R}^{n \times m}$.
The controllability matrix ( assuming a linear time invaraint system) can be defined as
$$ C = \begin{bmatrix}B & AB & A^2B & \cdots & A^{n-1}B \end{bmatrix} $$
and the system $ ( A, B) $ is controllable if $ C $ has full rank $ rank(C) = n $.
The rank of a matrix is the maximum number of linearly independent rows or columns. If $ m > n $ this means that $ rank(B) \leq n$.
From your description we know that $A $ is full rank, $ rank(A) = n$.
The rank of the product of two matrices obeys
$$ rank(AB) \leq \min(rank(A),rank(B)) $$
Combining all of this means that your system is only controllable if $ B $ is full rank. One intuitive way to think of this is having enough control inputs to control each of your states. Having a surplus of control inputs ($ m > n$) does not improve controllability.
Finally, $m<n$ is possible because the matrix product $ AB$ is well defined as long as $ B $ has $ n $ rows.
Hope this helps.