Why must $\gamma^\mu$ matrices in Dirac equation be linearly independent? In Ashok Das' Lecture on Quantum Field Theory, pg 20, it was said that the Dirac equation is derived by considering the Einstein relation $$p^\mu p_\mu=m$$
as a matrix relation, namely an $n\times n$ identity matrix multiplying both sides.
It was then said that we can take the square root of $p^\mu p_\mu$ to be $\mathcal{P}:$
$$\mathcal{P}=\gamma^\mu p_\mu$$
where $\gamma^\mu$ are four linearly independent $n\times n $ matrices.
I understand the motivation to introduce the matrices $\gamma^\mu$ to make the Einstein relation a matrix equation. But why is there a need for the $\gamma^\mu$ matrics to be linearly independent?
 A: If they were not, you would lose information on some component of momentum. For example, suppose for the sake of argument that $\gamma^3 = \alpha \gamma^1$, for some constant $\alpha$. Then we'd be able to write
$\gamma^0 p_0 + \gamma^1 p_1 + \gamma^2 p_2 + \gamma^3 p_3 = \gamma^0 p_0 + \gamma^1 (p_1 + \alpha p_3) + \gamma^2 p_2 = \gamma^0 p_0 + \gamma^2 p_2 + \gamma^3 (\frac{1}{\alpha} p_1 + p_3) = \mathcal{P}$
and $\mathcal{P}$ would keep no information of whether the particle is moving in the $x$ direction or in the $z$ direction, since they are getting mixed. We lose information about the system. If we want $\mathcal{P}$ to "store" all information that was available in $p^\mu$, then the $\gamma^\mu$ better be linearly independent.
A: Dirac matrices
with
\begin{align*}
   &\left({\mathbf{p}^\mu\,\mathbf{p}_\mu-m^2\,c^2}\right)\\&\overset{\text{Ansatz}}{=}\left({\beta^\kappa\,\mathbf{p}_\kappa+m\,c}\right)
   \left({\mathbf\gamma^\lambda\,\mathbf{p}_\lambda-m\,c}\right)\\&=
   \beta^\kappa\,\mathbf\gamma^\lambda\,\mathbf{p}_\kappa\,\mathbf{p}_\lambda+
   m\,c\left({\mathbf\gamma^\kappa-\beta^\kappa}\right)\,\mathbf{p}_\kappa-m^2\,c^2\tag 1
\end{align*}
\begin{align*}
&\Rightarrow\quad
   \beta^\kappa=\mathbf\gamma^\kappa
\end{align*}
hence equation (1)
\begin{gather*}
   \mathbf{p}^\mu\,\mathbf{p}_\mu= \beta^\kappa\,\mathbf\gamma^\lambda\,\mathbf{p}_\kappa\,\mathbf{p}_\lambda
   \quad\text{expanded }\\\\
   \left({\mathbf{p}^0}\right)^2- \left({\mathbf{p}^1}\right)^2- \left({\mathbf{p}^2}\right)^2- \left({\mathbf{p}^3}\right)^2\\=
   \left({\mathbf\gamma^0}\right)^2 \left({\mathbf{p}^0}\right)^2+\left({\mathbf\gamma^1}\right)^2 \left({\mathbf{p}^1}\right)^2+\left({\mathbf\gamma^2}\right)^2 \left({\mathbf{p}^2}\right)^2+\left({\mathbf\gamma^3}\right)^2 \left({\mathbf{p}^3}\right)^2\\
   +\left({\mathbf\gamma^0\,\mathbf\gamma^1+\mathbf\gamma^1\,\mathbf\gamma^0}\right)\mathbf{p}_0\,\mathbf{p}_1
   +\left({\mathbf\gamma^0\,\mathbf\gamma^2+\mathbf\gamma^2\,\mathbf\gamma^0}\right)\mathbf{p}_0\,\mathbf{p}_2\\
   +\left({\mathbf\gamma^0\,\mathbf\gamma^3+\mathbf\gamma^3\,\mathbf\gamma^0}\right)\mathbf{p}_0\,\mathbf{p}_3
   +\left({\mathbf\gamma^1\,\mathbf\gamma^2+\mathbf\gamma^2\,\mathbf\gamma^1}\right)\mathbf{p}_1\,\mathbf{p}_2\\
   +\left({\mathbf\gamma^1\,\mathbf\gamma^3+\mathbf\gamma^3\,\mathbf\gamma^1}\right)\mathbf{p}_1\,\mathbf{p}_3
   +\left({\mathbf\gamma^2\,\mathbf\gamma^3+\mathbf\gamma^3\,\mathbf\gamma^2}\right)\mathbf{p}_2\,\mathbf{p}_3
\end{gather*}
solution
\begin{align*}
 &\left({\mathbf\gamma^0}\right)^2=\mathbf 1\\
 &\left({\mathbf\gamma^i}\right)^2=-\mathbf 1\,,i=1\ldots 3\quad \text{and}\\
 &\mathbf\gamma^\mu\,\mathbf\gamma^\nu+\mathbf\gamma^\nu\,\mathbf\gamma^\mu=0 \quad,\mu \ne \nu
 \end{align*}
the solution is unique hence the matrices $~\mathbf\gamma^\mu~$ can't be linear dependent!
you obtain
\begin{align*}
   &\left({\mathbf{p}^\mu\,\mathbf{p}_\mu-m^2\,c^2}\right)=\left({\mathbf\gamma^\kappa\,\mathbf{p}_\kappa+m\,c}\right)
   \left({\mathbf\gamma^\lambda\,\mathbf{p}_\lambda-m\,c}\right)=0 \\
 \end{align*}
\begin{align*}
   &\mathbf\gamma^\mu\,\mathbf{p}_\mu-m\,c=0\quad \Rightarrow\\ \\ &\boxed{i\hbar\mathbf\gamma^\mu\partial_\mu\psi-m\,c\,\psi=0 }
 \end{align*}
Dirac- Matrices
\begin{align*}
&\mathbf\gamma^0=\begin{bmatrix}
      \mathbf{1} & 0 \\
      0 & -\mathbf{1} \\
    \end{bmatrix}\\
    &\mathbf\gamma^i=
    \begin{bmatrix}
      0 & \mathbf\sigma_i \\
      \mathbf\sigma_i & 0 \\
    \end{bmatrix}
\end{align*}
where  $~\mathbf\sigma_i$ are the  Pauli Matrices.
A: In the Dirac representation,the linear independence means   $$\sum_{i=1}^{4} \alpha_{i} \gamma^{i}=0 \Rightarrow \forall i\;, \alpha_{i}=0 $$
If there is a $ \;\alpha_{i}\neq 0 $ for which $\;\sum_{i=1}^{4} \alpha_{i} \gamma^{i}=0\; $ is that we have indirectly * a dimension which depends on the others, which contradicts the very essence of relativity. (4D 'representation' of the universe).
$*\;\;P=(E/c,\vec{p})=mV=m\frac{dX}{dt_{0}}=\gamma m(c,\frac{\vec{x}}{dt})\;\;,\;\;dt=\gamma dt_{0}$
