Is it true that there is a basis for R^(3x3) consisting only of nonsingular matrices.

I know that there should be 9 basis elements, but the only ones I can think of are the standard basis vectors, and I know that they are all singular.

Any ideas about how to attack this problem?

Question

Is it true that there is a basis for R^(3x3) consisting only of nonsingular matrices.

I know that there should be 9 basis elements, but the only ones I can think of are the standard basis vectors, and I know that they are all singular. 

Any ideas about how to attack this problem?

anonymous · Answer

How so? Isn't there a theorem that states that there must be mxm for R^(m x m) http://tutorial.math.lamar.edu/Classes/LinAlg/Basis.aspx Do a search for This follows from the natural extension of the previous part . That will take you to the theorem that I am speaking about

kinggeorge · Answer

I apologize, I was thinking it was 3-dimensional, not 3x3-dimensional.

anonymous · Answer

I understand basis and vectors in 3 dimensions, but once you slap on that R^(nxn) I'm totally lost!

kinggeorge · Answer

Same here.

anonymous · Answer

I know that if we want to prove this, we need to show that the span of the potential basis is the vector space, and that the elements are linearly independent, and also that the dimension of B is the correct number. My professor said if we can prove 2, the third is a consequence. So, we already know that we need 9 vectors, but now I just don't know how to prove the other ones!

anonymous · Answer

This apparently has the answer, but it is very confusing! http://www.cliffsnotes.com/study_guide/More-Vector-Spaces-Isomorphism.topicArticleId-20807,articleId-20795.html