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

Let $\struct {\map {\mathcal M_R} n, +, \times}$ denote the ring of square matrices of order $n$ over $R$.

The unit matrix:

$\mathbf I_n = \sqbrk a_n: a_{i j} = \delta_{i j}$

satisfies:

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

That is, the identity $\mathbf I_n$ for (conventional) matrix multiplication over $\struct {\map {\mathcal M_R} n, +, \times}$ 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 $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.

Let $\map {\mathcal M_R} {m, n}$ be the $m \times n$ matrix space over $R$.

Let $\mathbf A \in \map {\mathcal M_R} {m, n}$.

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

### Lemma: Right Identity

Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.

Let $\map {\mathcal M_R} {m, n}$ be the $m \times n$ matrix space over $R$.

Let $\mathbf A \in \map {\mathcal M_R} {m, n}$.

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

## Proof

### Proof of Lemma: Left Identity

Let $\sqbrk a_{m n} \in \map {\mathcal M_R} {m, n}$.

Let $\sqbrk b_{m n} = \mathbf I_m \sqbrk a_{m n}$.

Then:

 $\displaystyle \forall i \in \closedint 1 m, j \in \closedint 1 n \ \$ $\displaystyle b_{i j}$ $=$ $\displaystyle \sum_{k \mathop = 1}^m \delta_{i k} \circ a_{k j}$ $\displaystyle$ $=$ $\displaystyle a_{i j}$

Thus $\sqbrk b_{m n} = \sqbrk a_{m n}$ and $\mathbf I_n$ is shown to be a left identity.

$\Box$

### Proof of Lemma: Right Identity

Let $\sqbrk a_{m n} \in \map {\mathcal M_R} {m, n}$.

Let $\sqbrk b_{m n} = \sqbrk a_{m n} \mathbf I_n$.

Then:

 $\displaystyle \forall i \in \closedint 1 m, j \in \closedint 1 n: \ \$ $\displaystyle b_{i j}$ $=$ $\displaystyle \sum_{k \mathop = 1}^n a_{i k} \circ \delta_{k j}$ $\displaystyle$ $=$ $\displaystyle a_{i j}$

Thus $\sqbrk b_{m n} = \sqbrk a_{m n}$ and $\mathbf I_n$ is shown to be a right identity.

Hence $\mathbf I_n$ is an identity.

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

$\blacksquare$