Triangular Matrices forms Subring of Square Matrices

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Theorem

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

Let $\map {\MM_R} n$ be the order $n$ square matrix space over a ring $R$.

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


Let $\map {U_R} n$ be the set of upper triangular matrices of order $n$ over $R$.

Then $\map {U_R} n$ forms a subring of $\struct {\map {\MM_R} n, +, \times}$.


Similarly, let $\map {L_R} n$ be the set of lower triangular matrices of order $n$ over $R$.

Then $\map {L_R} n$ forms a subring of $\struct {\map {\MM_R} n, +, \times}$.


Proof

From Negative of Triangular Matrix, if $\mathbf B \in \map {U_R} n$ then $-\mathbf B \in \map {U_R} n$.

Then from Sum of Triangular Matrices, if $\mathbf A, -\mathbf B \in \map {U_R} n$ then $\mathbf A + \paren {-\mathbf B} \in \map {U_R} n$.

From Product of Triangular Matrices, if $\mathbf A, \mathbf B \in \map {U_R} n$ then $\mathbf A \mathbf B \in \map {U_R} n$.

The result follows from the Subring Test.


The same argument can be applied to matrices in $\map {L_R} n$.

$\blacksquare$


Sources

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