The defining property for the gamma matrices to generate a Clifford algebra is the anticommutation relation
$\displaystyle\{ \gamma^\mu, \gamma^\nu \} = \gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 \eta^{\mu \nu} I_4 $
where $\{ , \}$ is the anticommutator, $\eta^{\mu \nu}$ is the Minkowski metric with signature (+ − − −) and $I_4$ is the 4 × 4 identity matrix.
This makes perfect sense if $\eta^{\mu \nu}$ is a scalar that is equal to zero when $\mu \neq \nu$ and is either +1 or -1 when they are the same. This would make sense because squaring a gamma matrix results in either $I_4$ or $-I_4$
But everything I read online says that $\eta^{\mu \nu}$ is a tensor
$\eta^{\mu \nu} = \pm \begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix},$
Just for reference the 4 gamma matrices are:
$\begin{align} \gamma^0 &= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix},\quad& \gamma^1 &= \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{pmatrix} \\ \gamma^2 &= \begin{pmatrix} 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ 0 & i & 0 & 0 \\ -i & 0 & 0 & 0 \end{pmatrix},\quad& \gamma^3 &= \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}. \end{align}$
$\eta= \pm \begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix},$
but $\eta^{\mu \nu}$ is just one component of $\eta$