What is the correct name for those matrices where $M{i,j}$ = 1 if i = j, 0 o.w.?

Is there a name for matrices of this form?

$T\in \mathbb{R}^{m\times n}, T_{i,j}= \begin{array}{cc} \{ \begin{array}{cc} 1 & i=j \\ 0 & \text{True} \\ \end{array} \\ \end{array}$

Such as these:

$\begin{array}{cccc} \left( \begin{array}{c} 1 \\ \end{array} \right) & \left( \begin{array}{cc} 1 & 0 \\ \end{array} \right) & \left( \begin{array}{ccc} 1 & 0 & 0 \\ \end{array} \right) & \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ \end{array} \right) \\ \left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right) & \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right) & \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array} \right) & \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{array} \right) \\ \left( \begin{array}{c} 1 \\ 0 \\ 0 \\ \end{array} \right) & \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ \end{array} \right) & \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) & \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \end{array} \right) \\ \left( \begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \\ \end{array} \right) & \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ 0 & 0 \\ \end{array} \right) & \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{array} \right) & \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) \\ \end{array}$

• Since the entries are $T_{i,j}=\delta_{i,j}$ one could maybe call such a matrix a "Kronecker delta matrix", although I don't believe this is a popular or commonly used naming of such a matrix. – Dave Sep 8 '18 at 1:05
• There is no standard terminology, so whatever term you use you will have to explain it. I might call them "rectangular identity matrices" , taking care to explicitly define the term. – kimchi lover Sep 8 '18 at 1:46
• I like the Kronecker Delta Matrix, and yes I will keep in mind it is non-standard. Can I call the question answered? – Charles Gillingham Sep 8 '18 at 22:26