is a one-by-one-matrix just a number (scalar)? I was wondering. Clearly, we cannot multiply a (1x1)-matrix with a (4x3)-matrix; However, we can multiply a scalar with a matrix. This suggests a difference.

On the other hand, I was, for example, in an econometrics lecture today, where we had for a (Tx1)-vector $\underline{û}=\left(
\begin{array}{c}
û_1\\
\vdots\\
û_T\end{array}\right)$:


  $S_{ûû}:= \sum_{i=1}^T û_i^2$ shall be minimized. We see see that
  $S_{ûû}=\underline{û}^T\underline{û}$.


Well, formally, shouldn't it be $(S_{ûû})=\underline{û}^T\underline{û}$ or $S_{ûû}=\det(\underline{û}^T\underline{û})$, to ensure that we stay in the space of matrices and not suddenly go to the space of scalars? So here, the professor (physicist) not only treats $\underline{û}^T\underline{û}$ like a scalar, but also calls it a scalar. Is this formally legit or a wrong simplification (though it does not seem to have any impact, and surely makes life easier)?
It's just a scalar in the sense that the ring of $1\times 1$ matrices over a field $K$ is isomorphic to $K$ (by the map $[x]\mapsto x$), but, as you observed, when you're considering the interaction of matrices of different sizes, then you have to treat them differently.
No, a one-by-one matrix is not a scalar. 

Let's assume you want to multiply a one-by-one matrix $[c]$ with a 3x3 matrix $A$. To do this realize, that you can always extend a matrix to fit the dimensions by adding rows and columns of zeros. So you get
$$
\begin{bmatrix} c \end{bmatrix} \begin{bmatrix} a_{11} &amp; a_{12} &amp;   a_{13} \\ a_{21} &amp; a_{22} &amp;   a_{23} \\ a_{31} &amp; a_{32} &amp;   a_{33}  \end{bmatrix} = \begin{bmatrix} c &amp; 0 &amp;   0 \\ 0 &amp; 0 &amp;  0 \\ 0 &amp; 0 &amp;   0  \end{bmatrix} \begin{bmatrix} a_{11} &amp; a_{12} &amp;   a_{13} \\ a_{21} &amp; a_{22} &amp;   a_{23} \\ a_{31} &amp; a_{32} &amp;   a_{33}  \end{bmatrix} = \begin{bmatrix} ca_{11} &amp; ca_{12} &amp;   ca_{13} \\ 0 &amp; 0 &amp;   0 \\ 0 &amp; 0 &amp;   0  \end{bmatrix}
$$
But multiplying a scalar $c$ with $A$ is different. The scalar acts like a square diagonal matrix. So you get
$$
c  \begin{bmatrix} a_{11} &amp; a_{12} &amp;   a_{13} \\ a_{21} &amp; a_{22} &amp;   a_{23} \\ a_{31} &amp; a_{32} &amp;   a_{33}  \end{bmatrix} = \begin{bmatrix} c &amp; 0 &amp;   0 \\ 0 &amp; c &amp;  0 \\ 0 &amp; 0 &amp;   c  \end{bmatrix} \begin{bmatrix} a_{11} &amp; a_{12} &amp;   a_{13} \\ a_{21} &amp; a_{22} &amp;   a_{23} \\ a_{31} &amp; a_{32} &amp;   a_{33}  \end{bmatrix} = \begin{bmatrix} ca_{11} &amp; ca_{12} &amp;   ca_{13} \\ ca_{21} &amp; ca_{22} &amp;   ca_{23} \\ ca_{31} &amp; ca_{32} &amp;   ca_{33}  \end{bmatrix}
$$

From this we see: A one-by-one matrix $[c]$  always acts like a matrix with a single eigenvalue $c$. A scalar $c$ on the other hand is a shortcut for a n-dimensional matrix with $n$ equal eigenvalues $c$.