The Subring of n x n Upper and Lower Triangular Matrices

# 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)
\begin{align} \quad A + (-A) = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ 0 & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & a_{nn} \end{bmatrix} + \begin{bmatrix} -a_{11} & -a_{12} & \cdots & -a_{1n}\\ 0 & -a_{22} & \cdots & -a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & -a_{nn} \end{bmatrix} = \begin{bmatrix} a_{11} -a_{11} & a_{12} -a_{12} & \cdots & a_{1n} -a_{1n}\\ 0 & a_{22} -a_{22} & \cdots & a_{2n} -a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & a_{nn} -a_{nn} \end{bmatrix} = \begin{bmatrix} 0 & 0 & \cdots & 0\\ 0 & 0 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & 0 \end{bmatrix} \end{align}

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)
\begin{align} \quad (AB)_{ij} = \sum_{k=1}^{n} a_{ik}b_{kj} = a_{i1}b_{1j} + a_{i2}b_{2j} + ... + a_{in}b_{nj} \end{align}

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)
\begin{align} \quad (AB)_{ij} = \underbrace{a_{i1}b_{1j} + a_{i2}b_{2j} + ... + a_{ij}b_{jj} + ...}_{=0 \: \mathrm{since} \: a_{ij} = 0 \: \mathrm{for} \: i > j} + \underbrace{a_{ii}b_{ij} + ... + a_{in}b_{nj}}_{=0 \: \mathrm{since} \: b_{ij} = 0 \: \mathrm{for} \: i > j} = 0 \end{align}

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.