Do the four gamma matrices form a basis for the set of matrices $GL(4,\mathbb{C})$? I was actually trying to evaluate a term like $\gamma^0 M^\dagger \gamma^0$ in a representation independent way, where $M, M^\dagger$ are $4\times 4$ matrices.
Linear Algebra – Do Gamma Matrices Form a Basis? How to Understand Dirac and Clifford Algebra
clifford-algebradirac-matriceslinear algebra
Best Answer
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
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
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.