As the title says, I'm wondering whether there exists a basis for the set of $2\times 2$ matrices (with entries from the real numbers) such that all basis elements are invertible.
I have a gut feeling that it is false, but don't know how to prove it. I know that for a matrix to be invertible, it must be row equivalent to the identity matrix and I think I may be able to use this in the proof, but I don't know how.
Thanks in advance for any help,
Jack
Best Answer
Even without finding such a basis, you can see that singular matrices form a hypersurface in $ M_{n \times n}(F) $ given by the null set of the determinant map $ \det : M_{n \times n}(F) \to F $. When $ F = \mathbb R $, for instance, the set of all singular matrices is a closed subset of $ M_{n \times n}(F) $ (it is the preimage of $ \{ 0 \} $, which is closed, under the continuous determinant map) which is not all of the space, therefore there is an open ball lying outside of this set. As you can prove, we can then find a basis lying in this open ball, hence consisting of invertible matrices.
Explicit counterexamples have been given in other answers, so I will not mention any here.