Answer to Question #165250 in Abstract Algebra for K

Question

Prove or disprove: the set of diagonal n × n matrices with no zeroes on the diagonal is a subgroup of

GL(n,R)

Answer 1

$Solution: ~ Consider, the ~set~S~of ~the ~nXn~diagonal ~matrices ~ with ~ real ~ entries~ \\with ~ no ~ zeros ~ on ~ the ~diagoal~. \\Given ~ A=[a_{ij}]~ in ~S, Since ~ the~ determinant~ of~ a ~ diagonal ~ matrix ~ is ~ the ~ product ~ \\~ of ~ its ~diagonal ~ entries~ and ~ a_{ii}~\neq 0 ~for ~ all ~ 1\leq i \leq n~ \\ we ~have ~have ~that ~ det(A)=a_{11}a_{22}....a_{nn} \neq 0~which ~ gives ~ us ~that ~ A \in GL(n, R) . \\Hence ~S ~ \subset ~GL(n, R) . \\Given~ A=[a_{ij}]~ and~ B=[b_{ij}]~~ in ~S ~ we ~have ~that~ A ~and ~B~ are ~ both~ diagonal ~ \\matrices~ with ~ real ~ entries~ with ~ no ~zeros~ on ~ the ~ diagonal ~ which ~ gives~ us ~ that \\AB=[a_{ij}b_{ij}]. \\Since~ a_{ii} \neq 0 ~and~ b_{ii} \neq0~ for ~ all~ 1 \leq i \leq n, We ~ have ~ that ~ a_{ii}b_{ii} \neq0~ for ~ all~ 1 \leq i \leq n ~ and ~ therefore ~ AB \in S \\Since ~ I_n ~ is ~ a ~ diagonal ~ matrix ~ with~ no ~zeros~ on~ its ~ diagonal ~ we ~ have ~ that ~ I_n \in S. \\Given~ A=[a_{ij}]~in ~ S~ we ~ have ~ that ~ B=A^{-1}~ is ~ also ~ a ~ diagonal ~ matrix ~ with ~ real ~ \\entries~such~ that ~ b_{ii}=a_{ii}^{-1}~ for ~ all ~1 \leq i \leq n.Thus ~ since ~ a_{ii} \neq 0~for ~ all~1 \leq i \leq n,\\we ~ have ~ that ~ b_{ii}\neq 0 ~for~ all~ 1 \leq i \leq n~ and ~ therefore ~ A^{-1} \in S. \\Hence ~S ~ is ~ a ~ subgroup ~ of ~ GL(n,R). \\Hence~the~ set ~ of ~the ~diagonal ~nXn~matrices ~ with ~ real ~ entries~ \\with ~ no ~ zeros ~ on ~ the ~diagoal~is ~ a ~subgroup~ of~GL(n,R). \\Hence ~ proved.$