Do gamma matrices form a basis?

As previous answers have correctly noted gamma matrices do not forma a basis of $M(4,\mathbb{C})$. Nevertheless you can construct one from them in the following way

• 1 the identity matrix $\mathbb{1}$
• 4 matrices $\gamma^\mu$
• 6 matrices $\sigma^{\mu\nu}=\gamma^{[\mu}\gamma^{\nu]}$
• 4 matrices $\sigma^{\mu\nu\rho}=\gamma^{[\mu}\gamma^{\nu}\gamma^{\rho]}$
• 1 matrix $\sigma^{\mu\nu\rho\delta}=\gamma^{[\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\delta]}=i\epsilon^{\mu\nu\rho\delta}\gamma^5$

these 16 matrices form the basis that we were looking for.

Furthermore they are used to construct the spinor bilinears multiplying by $\bar{\psi}$ on the left and $\psi$ on the right, which transform in the Lorentz indices as follows

• $\bar{\psi}\psi$ scalar
• $\bar{\psi}\gamma^\mu\psi$ vector
• $\bar{\psi}\sigma^{\mu\nu}\psi$ 2nd rank (antisymmetric) tensor
• $\bar{\psi}\sigma^{\mu\nu\rho}\psi$ pseudovector
• $\bar{\psi}\gamma^5\psi$ pseudoscalar

the fact that they form a basis of of $M(4,\mathbb{C})$ is very important because these are then the only independent spinor bilinears (i.e $\bar{\psi}M\psi$) that can be constructed, any other can be expressed a linear combination of these. A different issue is if it would make any sense to sum any of these since they are different types of tensors under Lorentz group transformations.

To complement V. Moretti's excellent answer, it's worth emphasizing that the dimension of the four-by-four complex matrices $\mathbb C^{4\times 4}$, when seen as a vector space over $\mathbb C$, is $4\!\times\!4=16$. As such, a set of four matrices (i.e. vectors in $\mathbb C^{4\times 4}$) can never be a basis for it.

It's also worth saying that the general linear group $\text{GL}(4,\mathbb C)\subset\mathbb C^{4\times 4}$, i.e. the four-by-four matrices with nonzero determinant, is not a vector space (for one, it doesn't have a zero), and it is therefore misleading to speak of a basis for it. That said, it is still possible to ask for a minimal set of vectors (i.e. matrices) whose span will contain $\text{GL}(4,\mathbb C)$; this turns out to require a full basis of $\mathbb C^{4\times 4}$ because the matrices you 'skip', $\mathbb C^{4\times 4}\setminus\text{GL}(4,\mathbb C)$, have measure zero, so $\text{GL}(4,\mathbb C)$ is a complex manifold of dimension 16 and requires that many parameters to describe.

No they do not, due to dimensional reasons, but they are generators of the algebra. That is, $I$ and the products of $\gamma^a$ (products of one, two, three and four matrices) form such a basis.

NOTE ADDED. As Emilio Pisanty correctly remarked (also making some further interesting comments) $GL(4, \mathbb C)$ is not a linear space so questions about bases of it are inappropriate. In fact I implicitly interpreted that $GL(4,\mathbb C)$ as $M(4, \mathbb C)$, the complex algebra of $4\times 4$ complex matrices which, by definition, is also a complex vector space.