Unit Matrix is Identity for Matrix Multiplication/Left

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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 $m$.

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:

\(\ds \forall i \in \closedint 1 m, j \in \closedint 1 n: \, \) \(\ds b_{i j}\) \(=\) \(\ds \sum_{k \mathop = 1}^m \delta_{i k} a_{k j}\) where $\delta_{i k}$ is the Kronecker delta: $\delta_{i k} = 1_R$ when $i = k$ otherwise $0_R$
\(\ds \) \(=\) \(\ds a_{i j}\)


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

$\blacksquare$