TITLE: Does there exist a basis for the set of $2\times 2$ matrices such that all basis elements are invertible?
QUESTION [11 upvotes]: 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

REPLY [6 votes]: I just want to point out that, in order to prove that the answer to a question of the form

Does there exist a basis for $M_n(\mathbb{R})$ consisting of [matrices of some special form]?

is "yes", you only actually need to show that the special matrices span all of $M_n(\mathbb{R})$, since any spanning set can be reduced to a basis. 
It's usually a lot easier to check that some special set matrices are spanning than it is to explicitly describe a basis. Let's look at the case you asked about---invertible matrices. Let $A$ be any matrix. Take $\lambda$ to be a nonzero number which isn't an eigenvalue of $A$ (which can be done because $A$ can have at most $n$ eigenvalues). Then we have
$$A = \lambda I + (A-\lambda I)$$
where $\lambda I$ and $A-\lambda I$ are both invertible. Since every matrix can be written as a sum of two invertibles, the invertibles are spanning, and can, in principal, be reduced to a basis.