Unit Matrix is Unity of Ring of Square Matrices

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

Let $\mathcal M_R \left({n}\right)$ be the $n \times n$ matrix space over $R$.

The identity matrix:


 * $\mathbf I_n = \left[{a}\right]_{n}: a_{i j} = \delta_{i j}$

satisfies:


 * $\forall \mathbf A \in \mathcal M_R \left({n}\right): \mathbf A\mathbf I_n = \mathbf I_n = \mathbf I_n \mathbf A$

That is, the identity $\mathbf I_n$ for (conventional) matrix multiplication over $\left({\mathcal M_R \left({n}\right), +, \times}\right)$ is a square matrix where every element on the diagonal is equal to $1_R$, and which is $0_R$ elsewhere.

Lemma: Left Identity
Let $\left({R, +, \circ}\right)$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.

Let $\mathcal M_R \left({m, n}\right)$ be the $m \times n$ matrix space over $R$.

Let $\mathbf A \in \mathcal M_R \left({m, n}\right)$.

Then $\mathbf I_m \mathbf A = \mathbf A$.

Lemma: Right Identity
Let $\left({R, +, \circ}\right)$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.

Let $\mathcal M_R \left({m, n}\right)$ be the $m \times n$ matrix space over $R$.

Let $\mathbf A \in \mathcal M_R \left({m, n}\right)$.

Then $\mathbf A \mathbf I_n = \mathbf A$.

Proof of Lemma: Left Identity
Let $\left[{a}\right]_{m n} \in \mathcal M_R \left({m, n}\right)$.

Let $\left[{b}\right]_{m n} = \mathbf I_m \left[{a}\right]_{m n}$. Then:

Thus $\left[{b}\right]_{m n} = \left[{a}\right]_{m n}$ and $\mathbf I_n$ is shown to be a left identity

Proof of Lemma: Right Identity
Let $\left[{a}\right]_{m n} \in \mathcal M_R \left({m, n}\right)$.

Let $\left[{b}\right]_{m n} = \left[{a}\right]_{m n} \mathbf I_n$. Then:

Thus $\left[{b}\right]_{m n} = \left[{a}\right]_{m n}$ and $\mathbf I_n$ is shown to be a right identity

Hence $\mathbf I_n$ is a two-sided identity.

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