Unit Matrix is Identity for Matrix Multiplication

Theorem

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

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

Let $\map {\MM_R} n$ denote the metric space of square matrices of order $n$ over $R$.

Let $\mathbf I_n$ denote the unit matrix of order $n$:

Then:

$\forall \mathbf A \in \map {\MM_R} n: \mathbf A \mathbf I_n = \mathbf A = \mathbf I_n \mathbf A$

That is, the unit matrix $\mathbf I_n$ is the identity element for (conventional) matrix multiplication over $\map {\MM_R} n$.

Proof

Lemma: Left Identity

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:

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

$\Box$

Lemma: Right Identity

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

Let $I_n$ denote the unit matrix of order $n$.

Then:

$\forall \mathbf A \in \map {\MM_R} {m, n}: \mathbf A \mathbf I_n = \mathbf A$

$\Box$

Thus: $\mathbf A \mathbf I_n = \mathbf A = \mathbf I_n \mathbf A$

Hence, by definition, $\mathbf I_n$ is an identity element for (conventional) matrix multiplication over $\map {\MM_R} n$.

That $\mathbf I_n$ is the identity element follows from Identity is Unique.

$\blacksquare$

Sources

• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $1$: Definitions and Examples: Example $1.7$
• 1998: Richard Kaye and Robert Wilson: Linear Algebra ... (previous) ... (next): Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.2$ Addition and multiplication of matrices: $10$