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:


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

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

That is, the identity $$\mathbf{I_n}$$ for Matrix multiplication (conventional) 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.