Missing identity element in the Clifford relation While studying the Dirac equation, $$\left(i\gamma^{\mu} \partial_{\mu} - m\right)\psi = 0.$$ I have been finding difficulty understanding the following summarisation of the algebra that the $\gamma$-matrices follow, $$\{\gamma^{\mu},\gamma^{\nu}\} = 2 \eta^{\mu \nu}$$ Where $\eta^{\mu \nu}$ is the inverse Minkowski metric, and the curly brackets denote an anticommutator.
Now, $\{\gamma^{0},\gamma^{0} \} = 2I$, so does this mean $\eta^{00} = I$? What does it mean (if anything) to substitute numbers for $\mu$ and $\nu$ here? My main question is, how does this "summary" relate to $\left(\gamma^{0}\right)^2 = I$, $(\gamma^k)^2 = -I$ for $k = 1,2,3$, and $\{\gamma^\mu,\gamma^\nu \} = 0$ for $\mu \neq \nu$?
Something tells me it relates to the matrix elements of $\eta^{\mu \nu}$, as the off-diagonal elements are zero, and the top diagonal element is 1. So one can look at the entries of $\eta^{\mu \nu}$ to predict what the anticommutators of the $\gamma$-matrices will evaluate to. 
 A: It can be easier to write everything in terms of explicit indices,
$$\gamma^\mu_{\alpha \beta} \gamma^\nu_{\beta \delta} + \gamma^\nu_{\alpha \beta} \gamma^\mu_{\beta \delta} = 2 \eta^{\mu\nu} \delta_{\alpha \delta}.$$
There is a sum over $\beta$ on the left-hand side. When you plug in explicit values for $\alpha$, $\delta$, $\mu$, and $\nu$, both sides are just numbers. The indices $\mu$ and $\nu$ are Lorentz indices, describing the Lorentz transformation properties of both sides. The indices $\alpha$, $\beta$, and $\delta$ are spinor indices, which transform in a different way. It is a complete coincidence they both range from $1$ to $4$. 
You can see how this looks clunky, so we could just factor the spinor indices out,
$$(\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu)_{\alpha \delta} = 2 \eta^{\mu\nu} \delta_{\alpha \delta}.$$
Then the matrix multiplication on the left-hand side becomes implicit. We can go one step further and suppress the spinor indices entirely, giving
$$\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 \eta^{\mu\nu}.$$
The problem is that the left-hand side is still clearly a $4 \times 4$ matrix in spinor space, but the right-hand side doesn't seem to have any spinor indices at all. They're there, but there isn't a simple way to show it. You could write $\eta^{\mu\nu} 1_4$ on the right-hand side, but that invites some confusion as it looks like the $1_4$ multiplies $\eta^{\mu\nu}$. You can't indicate it's an identity matrix in spinor space. The usual abbreviated notation is not perfect, but it's one of the best options we have.
A: Let's try to summarize the characteristics of the $\gamma$-matrices and the metric tensor for flat Minkowski space. First our $\eta^{\mu\nu}$ can be represented by the matrix 
$$\eta^{\mu\nu} = \begin{pmatrix} 1&0&0&0 \newline 0&-1&0&0\newline 0&0&-1&0\newline 0&0&0&-1 \\  \end{pmatrix}.$$
You now interpret the $\eta^{\mu\nu}$ as the $\mu\nu$-component of that matrix. So it doesn't make sense to say that $\eta^{00} = I_4$ but you can conclude that the $00$-component of your metric tensor is identical to the $00$-component of $I_4$ ($I^{00} = 1$). 
If you look at $(\gamma^\mu)^2$ then you can show following identity 
$$2\eta^{\mu\mu}I_4=\{\gamma^\mu;\gamma^\mu\} = \gamma^\mu \gamma^\mu + \gamma^\mu \gamma^\mu = 2(\gamma^\mu)^2.$$
For the off-diagonal elements you will obtain similar that 
$$2\eta^{\mu\nu}I_4=\{\gamma^\mu;\gamma^\nu\} = \gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = \gamma^\mu\gamma^\nu - \gamma^\mu\gamma^\nu = 0 \cdot I_4.$$
A: FWIW, a similar issue arises in the CCR
$$ [\hat{q}^j, \hat{p}_k]~=~i\hbar ~\delta^j_k \hat{\bf 1} $$
of the Heisenberg algebra, where authors often do not write the identity operator $\hat{\bf 1}$ explicitly.
