Notation
$\gamma$ are gamma matrices [0].
Context
I am working through the proof of relativistic invariance of the Dirac equation that is presented by Bethe and Jackiw [1]. They write that
The most general homogeneous Lorentz transformation between two coordiante systems may be written as $$ {x'} = {a^\mu}_\nu \,x^\nu... $$ Defining $\gamma'^\lambda = {a^\lambda}_\mu\,\gamma^\mu $, we can verify with the help of $$ {a_\mu}^\nu {a^\mu}_\lambda = a^{\mu\nu} a_{\mu\lambda} = a^{\nu\mu} a_{\lambda\mu} = {\delta^\nu}_\lambda $$ that ${\gamma'}^\lambda$ satisfies \begin{align*} \gamma^\mu\gamma^\nu + \gamma^\nu\gamma^\mu = 2\,g^{\mu\nu} \\ \gamma_\mu\gamma_\nu + \gamma_\nu\gamma_\mu = 2\,g_{\mu\nu}. \end{align*}
So, I tried to verify this on my own and I did not succeed.
Incorrect Derivation
To show that I am not understanding this material it suffices to look at a simpler case.
\begin{align*} {\gamma'}^\lambda {\gamma'}^\lambda &= {\gamma'}^\mu {\gamma'}^\mu \\ &= \left[ {a^\lambda}_0 \gamma^0 + {a^\lambda}_1 \gamma^1 + {a^\lambda}_2 \gamma^2 + {a^\lambda}_3 \gamma^3 \right] \left[ {a^\lambda}_0 \gamma^0 + {a^\lambda}_1 \gamma^1 + {a^\lambda}_2 \gamma^2 + {a^\lambda}_3 \gamma^3 \right] \\ &= \left[ {a^\lambda}_0 {a^\lambda}_0 \gamma^0 \gamma^0 + {a^\lambda}_1 {a^\lambda}_1 \gamma^1 \gamma^1 + {a^\lambda}_2 {a^\lambda}_2 \gamma^2 \gamma^2 + {a^\lambda}_3 {a^\lambda}_3 \gamma^3 \gamma^3 \right] \\ &+ {a^\lambda}_0 {a^\lambda}_1 \left( \gamma^0 \gamma^1 + \gamma^1 \gamma^0 \right) + {a^\lambda}_0 {a^\lambda}_2 \left( \gamma^0 \gamma^2 + \gamma^2 \gamma^0 \right) + {a^\lambda}_0 {a^\lambda}_3 \left( \gamma^0 \gamma^3 + \gamma^3 \gamma^0 \right) \\ &+ {a^\lambda}_1 {a^\lambda}_2 \left( \gamma^1 \gamma^2 + \gamma^2 \gamma^1 \right) + {a^\lambda}_1 {a^\lambda}_3 \left( \gamma^1 \gamma^3 + \gamma^3 \gamma^1 \right) \\ &+ {a^\lambda}_2 {a^\lambda}_3 \left( \gamma^2 \gamma^3 + \gamma^3 \gamma^2 \right) \end{align*} I know that when $\mu\neq \nu$ the anticommutator $\left\{\gamma^\mu,\gamma^\nu\right\}$ reduces to $0$, where $0$ is the zero matrix. Hence \begin{align*} {\gamma'}^\lambda {\gamma'}^\lambda &= \left[ {a^\lambda}_0 {a^\lambda}_0 \gamma^0 \gamma^0 + {a^\lambda}_1 {a^\lambda}_1 \gamma^1 \gamma^1 + {a^\lambda}_2 {a^\lambda}_2 \gamma^2 \gamma^2 + {a^\lambda}_3 {a^\lambda}_3 \gamma^3 \gamma^3 \right] \\ &= \left[ {a^\lambda}_0 {a^\lambda}_0 g_{00} + {a^\lambda}_1 {a^\lambda}_1 g_{11} + {a^\lambda}_2 {a^\lambda}_2 g_{22} + {a^\lambda}_3 {a^\lambda}_3 g_{33} \right]I \\ &= \pm \left[ {a^\lambda}_0 {a^\lambda}_0 - {a^\lambda}_1 {a^\lambda}_1 - {a^\lambda}_2 {a^\lambda}_2 - {a^\lambda}_3 {a^\lambda}_3 \right]I \end{align*}
What I expected to obtain in the end was $\pm I$, where $I$ is the identity matrix. Yet, I do not obtain this result. If the Lorentz transformation matrix were symmetric, then I would obtain the identity matrix. Yet, I know that, in general, the Lorentz transformation matrix cannot be taken as symmetric.
Question(s)
Am I misunderstanding what is meant by the operation ${\gamma'}^\lambda {\gamma'}^\lambda $? Is it implied that the rightmost matrix is to be transposed? Otherwise, how do I go about verifying what there is to verify?
Bibliography
[0] https://en.wikipedia.org/wiki/Gamma_matrices
[1] Bethe and Jackiw, Intermediate Quantum Mechanics, pp. 354, 361.
