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 $R$.

Then the ring $\left({\mathcal M_R \left({n}\right), + \times}\right)$ has a unity:


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

where $\delta_{i j}$ is the Kronecker delta.

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 $0_R$ elsewhere.

This is called the identity matrix.

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
Putting the two results together from the following, the result is shown to hold.

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 the result holds.

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 the result holds.