Let $\eta$ be a pseudo Euclidian inner product of signature $(t,s)$ (i.e. in matrix notation $t$ $-1$'s down the diagonal, followed by $s$ $+1$'s down the diagonal, or rather in any orthonormal basis there are $t$ $e_i$'s which satisfy $\eta(e_i,e_i)=-1$, and $s$ $e_i$'s which satisfy $\eta(e_i,e_i)=+1$) on $\mathbb{R}^{t+s}$. Furthermore let $SO^+(t,s)$ be orthochronus special pseudo orthogonal group, which is the group of matrices with preserve $\eta$, the orientation on $\mathbb{R}^{t,s}$, and the orientation of any maximally negative definite subspace. We denote by $Cl(t,s)$ the Clifford algebra of $\mathbb{R}^{t+s}$ with $\eta$ as the symmetric bilinear form. Finally, one can show that the set:
$$\text{Spin}^+(t,s)=\{v_1\cdots v_{2n}: \eta(v_i,v_i)=\pm 1, \text{and an even number of $v_i$ have magnitude $-1$}\}$$
is the double cover of $SO^+(t,s)$. Note that the set above is a subset of the $CL(t,s)$, so the product is multiplication in $CL(t,s)$, and that there are an even amount of $v_i$ in the product which satisfy $\eta(v_i,v_i)=+1$. One can show that the Lie algebra of $\mathfrak{spin}^+(t,s)$ is given by the span of the set:
$$\{e_ie_j:1\leq i<j\leq s+t\}$$
Furthermore, the Lie algebra of $SO^+(t,s)$ is given by: $$\{X\in \text{Mat}_{n\times n}(\mathbb{R}): X^T\eta+\eta X=0\}$$
Let $\lambda:\text{Spin}^+(t,s)\rightarrow SO^+(t,s)$ be the double cover Lie group homorphism. Since $\lambda$ is surjective, and the kernel is finite, it follows that $\lambda_*:\mathfrak{spin}^+(t,s)\rightarrow \mathfrak{so}^+(t,s)$ is a Lie algebra isomorphism. Let $\{e_i\}$ be an $\eta$ orthonormal basis for $\mathbb{R}^{t+s}$, let $\{e^i\}$ be the $\eta$ dual frame. We define the coefficients $$T^i_j=e^i\otimes e_j$$ which is really just the $s+t$ by $s+t$ matrix of all zeroes, except a $1$ in the $i$th column of the $j$th row. A calculation shows that:
$$\lambda_*(e_ie_j)=2(\eta_{ii}T_j^i-\eta_{jj}T_i^j$$
Now, let $A\in \mathfrak{so}^+(t,s)$, then for some real coefficients $A^c_a$, we have that: $$A=A^c_aT^a_c$$ where Einstein summation convention is implied (going forward I will explicitly mention when I do not imply it). It follows that the coefficients satisfy:
$$A^c_a\eta_{cb}+A^{c}_b\eta{ca}=0$$
as this is just the property that $A^T\eta+\eta A=0$. Define coefficients $A_{ab}$ by: $$A_{ab}=A^c_a\eta_{cb}$$ and then trivially: $$A=A_{ab}\eta^{bc}T^a_c$$ where $A_{ab}=-{ba}$.
I have reason to believe that the inverse of $\lambda_*$ is given by:
$$(\lambda^{-1}_*)(A)=\frac{1}{8}A_{ac}[\eta^{ab}e_b,\eta^{cd}e_d]$$
As under the spinor representation, where $e_i$ gets mapped to the mathematical gamma matrix $\gamma_i$, this becomes:
$$\frac{1}{4}A_{ab}\gamma^{ab}$$ where:
$$\gamma^a=\eta^{ab}\gamma_b$$ and: $$\gamma^{ab}=\frac{1}{2}[\gamma^a,\gamma^b]$$
However, with:
$$A=A_{ab}\eta^{bc}T^a_c$$ we see that $\eta^{bc}=0$ unless $c =b$, hence if forgo Einstein summation convention we write: $$A=\sum_{a,c}A_{ac}\eta^{cc}T^a_c$$ Furthermore, $\eta$ is a diagonal matrix of $1$'s and $-1$'s we get that $\eta^{cc}=\eta_{cc}$, as $\eta^{cc}$ denote the components of the inverse matrix. It follows that we can rewrite this as: $$A=\sum_{a,c}A_{ac}\eta_{cc}T^a_c$$ Since $A_{ac}$ is antisymmetric, we can write this as: $$\begin{align} A_a^cT^c_a=&\frac{1}{2}\sum_{a,c}A_{ac}\eta_{cc}T^a_c+A_{ca}\eta_{aa}T_a^c\\ =&\frac{1}{4}\sum_{a,c}A_{ac}2\left(\eta_{cc}T^a_c-\eta_{aa}T_a^c\right) \end{align}$$ so $(\lambda_*)^{-1}$ is given by:
$$(\lambda^*)^{-1}(A)=\frac{1}{4}\sum_{ac}A_{ac}e_ae_c$$
Did I mess something up here? I'm really confused, the book I'm using says the result in terms of the mathematical gamma matrices come trivially from the formula that for any $z\in \mathfrak{spin}^+(t,s)$ we can recover $z$ from its image $\lambda_*(z)$ by:
$$z=\sum_{a<c}\eta(\lambda_*(z)e_a,e_l)\eta_{aa}\eta_{cc}e_ae_c$$
But $a)$, I don't see how to obtain this formula, and $b)$ I don't see how I can use to obtain $(\lambda^*)^{-1}(A)$ for some $A\in \mathfrak{so}^+(t,s)$. Any help would be greatly appreciated.