Unit Matrix is Identity for Matrix Multiplication/Left

Theorem
Let $R$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.

Let $m, n \in \Z_{>0}$ be a (strictly) positive integer.

Let $\map {\MM_R} {m, n}$ denote the $m \times n$ metric space over $R$.

Let $I_m$ denote the unit matrix of order $m$:
 * $\mathbf I_m = \sqbrk a_m: a_{i j} = \delta_{i j}$

Then:
 * $\forall \mathbf A \in \map {\MM_R} {m, n}: \mathbf I_m \mathbf A = \mathbf A$

Proof
Let $\sqbrk a_{m n} \in \map {\MM_R} {m, n}$.

Let $\sqbrk b_{m n} = \mathbf I_m \sqbrk a_{m n}$.

Then:

Thus $\sqbrk b_{m n} = \sqbrk a_{m n}$ and $\mathbf I_m$ is shown to be a left identity.