# The Subring of n x n Upper and Lower Triangular Matrices

Recall from the Subrings and Ring Extensions page that if $(R, +, *)$ is a ring, $S \subseteq R$, and $S$ forms a ring under the same $+$ and $*$ defined on $R$ then the ring $(S, +, *)$ is said to be a subring of $(R, +, *)$ and $(R, +, *)$ is said to be a ring extension of $(S, +, *)$.

We also saw that if $(R, +, *)$ is a ring where $0$ is the identity of $+$ and $1$ is the identity of $*$ then we can check if a subset $S \subseteq R$ is a ring by verifying the following ring axioms:

- $S$ is closed under $+$.

- For every element $a \in S$ there exists a $-a \in S$ such that $a + (-a) = 0$ and $(-a) + a = 0$.

- $S$ is closed under $*$.

- $1 \in S$.

Now recall from The Ring of n x n Matrices page that the set $M_{nn}$ of $n \times n$ matrices with real coefficients form a ring under the operation of standard matrix addition $+$ and standard matrix multiplication $*$.

Let $U_{nn} \subseteq M_{nn}$ be the set of all $n \times n$ upper triangular matrices with real coefficients and let $L_{nn} \subseteq M_{nn}$ be the set of all $n \times n$ lower triangular matrices with real coefficients. We will first show that $U_{nn}$ forms a subring under $+$ and $*$ by verifying the axioms above. Let $A, B, C \in U_{nn}$.

Clearly $U_{nn}$ is closed under addition since the sum of two upper triangular matrices $A$ and $B$ is necessarily an upper triangular matrix since we sum up corresponding matrix entries to form $A + B$. The entries below the main diagonal of $A$ and $B$ are zero and hence the entries below the main diagonal of $A + B$ are zero, so $(A + B) \in U_{nn}$.

For $A \in U_{nn}$ we can define $-A \in U_{nn}$ to be the upper triangular matrix entries on and above the main diagonal are the negations of the entries on and above the main diagonal of $A$. Then:

(1)Similarly $(-A) + A = 0_{n \times n}$ due to the commutativity inherited on $U_{nn}$ from $M_{nn}$.

Now consider the product $A * B = AB$ of upper triangular matrices and let $(AB)_{ij}$ be the entries in this product for $i, j \in \{1, 2, ..., n \}$. Then:

(2)Since $A$ and $B$ are upper triangular matrices we have that whenever $i > j$ that then $a_{ij} = 0$ and $b_{ij} = 0$. So if $i > j$ then:

(3)Therefore $A * B \in U_{nn}$.

Lastly, the identity matrix $I_n$ is an upper triangular matrix, so $I_n \in U_{nn}$.

Thus, $(U_{nn}, +, *)$ is a ring. It can be shown that $(L_{nn}, +, *)$ is also a ring using an analogous approach as above.