Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Determine whether the given set of invertible n x n matrices with real number entries is a subgroup of $GL(n,\mathbb{R})$

The n x n matrices with determinant 2

The key saidenter image description here

I understand that's how you compute determinants, but why did they pick out multiplication? Why couldn't they say for instance

"If detA = detB = 2, then we also see that detA + detB = 4 and it is not closed under addition".

share|improve this question
3  
Note that it is not true that $\det(AB)=\det A+\det B$, nor is it true that $\det(A+B)=\det A+\det B$, so the fact $\det A+\det B=4$ would be worthless, whereas $\det(A)\det(B)=4$ tells you $AB$ has determinant $4$ and hence $AB$ is not in the set. –  anon Oct 15 '12 at 5:28

2 Answers 2

up vote 2 down vote accepted

"Subgroup of $G$" means "subgroup under the operation inherited from $G$". The operation on the big group is multiplication, not addition.

share|improve this answer
    
So $gL(n,\mathbb{R})$ means addition then? –  jip Oct 15 '12 at 5:02
    
@jak: $GL(n,F)$ (capital letters) stands for the general linear group of $F^n$ ($F$ is the base field), which consists of invertible $n\times n$ matrices with the group operation being multiplication. In the context of Lie theory, $\mathfrak{gl}(V)$ (lowercase fraktur font) is the general linear algebra: it contains all endomorphisms of $V$, and is an algebra $-$ it has both addition and multiplication as two different binary operations. –  anon Oct 15 '12 at 5:17
    
In the context of matrices, "invertible" means "having a multiplicative inverse". So you've got the matrices that have a multiplicative inverse. That set is not closed under addition, so when we say it's a group, we must be talking about multiplication. I don't know if your lower-case $g$ is a typo or a genuine distinction; I've never seen any $gL(n,{\bf R})$. –  Gerry Myerson Oct 15 '12 at 6:35

It's because the group operation in $GL(n,\mathbb{R})$ is taken to be matrix multiplication so we want to see if the product of two matrices with determinant 2 still has determinant 2 when we are trying to determine whether the set of such matrices is a subgroup.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.