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.
$\blacksquare$