# Thread: Proving a set of matrices is a group?

1. ## Proving a set of matrices is a group?

I'm looking to prepare myself for all sorts of proofs which deal with sets of matrices as groups. Here's an example I'm having a tough time recalling how to complete.

Show that the set GL2(R) of all invertible 2x2 matrices with real
entries is a group under usual matrix multiplication. Is the group abelian?

Showing that the group is abelian will hopefully be easy once I prove the original statement. I should simply take two arbitrary matrices and multiply them, then swap them and multiply to view that I shall most likely have a different result (I don't believe matrices commute).

Thanks for any help!

2. Remember the four defining properties of a group: closure, associativity, identity element, and inverse:
1. Closure - For every pair of elements a and b, a*b must be an element of the group.
2. Associativity - For any three elements a, b, and c, the equality a*(b*c)=(a*b)*c must hold.
3. Identity element - There is a group element e, the identity element, such that a*e=e*a=a for any a in the group.
4. Inverse - For every group element a, there is an element b, the inverse, such that a*b=b*a=e, where e is the identity.

You just need to show that the set of all invertible 2x2 matrices with real
entries under matrix multiplication fits these. For example, the existence of the 2x2 identity matrix establishes #3, the matrices being invertible satisfies #4.

--Kevin C.

3. Originally Posted by TwistedOne151
Remember the four defining properties of a group: closure, associativity, identity element, and inverse:
1. Closure - For every pair of elements a and b, a*b must be an element of the group.
2. Associativity - For any three elements a, b, and c, the equality a*(b*c)=(a*b)*c must hold.
3. Identity element - There is a group element e, the identity element, such that a*e=e*a=a for any a in the group.
4. Inverse - For every group element a, there is an element b, the inverse, such that a*b=b*a=e, where e is the identity.

You just need to show that the set of all invertible 2x2 matrices with real
entries under matrix multiplication fits these. For example, the existence of the 2x2 identity matrix establishes #3, the matrices being invertible satisfies #4.

--Kevin C.
thanks for the response! should i be physically drawing up arbitrary matrices, with elements [a b, c d] and [e f, g h] for example, and multiplying them together to show that they give another element in the group?

thanks again

4. Originally Posted by charleschafsky
should i be physically drawing up arbitrary matrices, with elements [a b, c d] and [e f, g h] for example, and multiplying them together to show that they give another element in the group?
Probably not. What you need to establish is that the product of two invertible matrices is invertible. For that, use what you know about determinants.

5. Awesome thanks a lot. Here's what I've got so far.

Identity: By the invertible matrix theorem, all square invertible matrices are row equivalent to the identity matrix.

Inverse: By theorem, for all A, A exists in GL2(R), there exists a B, B exists in GL2(R), such that AB = BA = I.

Associativity was a huuuge waste of time. I took three arbitrary matrices and did the multiplication. I'm not sure if there's a quicker way to this.

Closure I was having a bit of trouble with. I realize that the determinant must not be 0 for the resultant matrix to be invertible. I assumed I'd end up with some sort of contradiction where the only way the final determinant could be 0 would be if certain values that could not be 0 in the original invertible matrices were 0, but unless I've messed up somewhere I just ended up with aedh + bgcf - afdg - bhec. This isn't very useful, or meaningful.

Thanks again for everybody's help

6. Originally Posted by charleschafsky
Awesome thanks a lot. Here's what I've got so far.

Identity: By the invertible matrix theorem, all square invertible matrices are row equivalent to the identity matrix. Not sure what that is about. The identity matrix is the identity for the group because AI = IA = A for all A in GL2(R)

Inverse: By theorem, for all A, A exists in GL2(R), there exists a B, B exists in GL2(R), such that AB = BA = I.

Associativity was a huuuge waste of time. I took three arbitrary matrices and did the multiplication. I'm not sure if there's a quicker way to this. I think you're probably allowed to quote the fact that matrix multiplication is always associative. Or is that something that you haven't come across before?

Closure I was having a bit of trouble with. I realize that the determinant must not be 0 for the resultant matrix to be invertible. I assumed I'd end up with some sort of contradiction where the only way the final determinant could be 0 would be if certain values that could not be 0 in the original invertible matrices were 0, but unless I've messed up somewhere I just ended up with aedh + bgcf - afdg - bhec. This isn't very useful, or meaningful. I was hoping that you would know the result that det(AB) = det(A)det(B). It's obvious from that that if det(A) and det(B) are nonzero then so is det (AB).
..

7. That last part entirely slipped my mind. Thanks a lot. You've been super helpful

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# set of nxn matrices is a group under multiplication

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