Deriving $\Lambda^i_{\,j}$ components of the Lorentz transformation matrix

I would like to see a general quick derivation that follows on exactly in the spirit of Weinberg.

Recall this involves considering a particle in a frame $\mathcal{O}$ in which the particle appears to have no velocity and also in a frame $\mathcal{O}'$ in which it appears to have velocity $$ \mathbf{v} = (\frac{dx'^i}{dt'}),$$ and then using $$dt' = \Lambda^0_{\,\,\,0} dt,$$ $$dx'^i = \Lambda^i_{\,\,\,0} dt,$$ along with the $00$ component of the identity $$\Lambda^{\alpha}_{\,\,\,\gamma} \Lambda^{\beta}_{\,\,\,\delta} \eta_{\alpha \beta} = \eta_{\gamma \delta}$$ to show that a Lorentz transformation $\Lambda^{\alpha}_{\,\,\,\beta}$ between these frames must satisfy $$\Lambda^0_{\,\,\,0} = \gamma$$ $$\Lambda^i_{\,\,\,0} = \gamma v^i,$$ while the remaining $\Lambda^{\alpha}_{\,\,\,\beta}$ components are not uniquely fixed as above since for any rotation $R^{\alpha}_{\,\,\,\beta}$ we have that both $\Lambda^{\alpha}_{\,\,\,\beta}$ and $\Lambda^{\alpha}_{\,\,\,\gamma} R^{\gamma}_{\,\,\,\beta}$ will transform from the frame where the particle appears with zero velocity to the frame where it appears to have velocity $\mathbf{v}$.

One way, which is similar to this, is to first consider the special case (boo) of a frame in which $v^1 = v, v^2 = 0, v^3 = 0$ and then try to write the results so that it easily applies for any $\mathbf{v}$. Thus, given $$ \Lambda^{\alpha}_{\,\,\,\beta} = \begin{bmatrix} \gamma & \gamma v & 0 & 0 \\ \gamma v & \Lambda^1_{\,\,\,1} & \Lambda^1_{\,\,\,2} & \Lambda^1_{\,\,\,3} \\ 0 & \Lambda^2_{\,\,\,1} & \Lambda^2_{\,\,\,2} & \Lambda^2_{\,\,\,3} \\ 0 & \Lambda^3_{\,\,\,1} & \Lambda^3_{\,\,\,2} & \Lambda^3_{\,\,\,3} \end{bmatrix} $$ use the fact that \begin{align} 1 &= \det(\Lambda^{\alpha}_{\,\,\,\beta} ) \\ &= \gamma \begin{bmatrix} \Lambda^1_{\,\,\,1} & \Lambda^1_{\,\,\,2} & \Lambda^1_{\,\,\,3} \\ \Lambda^2_{\,\,\,1} & \Lambda^2_{\,\,\,2} & \Lambda^2_{\,\,\,3} \\ \Lambda^3_{\,\,\,1} & \Lambda^3_{\,\,\,2} & \Lambda^3_{\,\,\,3} \end{bmatrix} - \gamma v \begin{bmatrix} \gamma v & 0 & 0 \\ \Lambda^2_{\,\,\,1} & \Lambda^2_{\,\,\,2} & \Lambda^2_{\,\,\,3} \\ \Lambda^3_{\,\,\,1} & \Lambda^3_{\,\,\,2} & \Lambda^3_{\,\,\,3} \end{bmatrix} \\ &= \gamma \begin{bmatrix} \Lambda^1_{\,\,\,1} & \Lambda^1_{\,\,\,2} & \Lambda^1_{\,\,\,3} \\ \Lambda^2_{\,\,\,1} & \Lambda^2_{\,\,\,2} & \Lambda^2_{\,\,\,3} \\ \Lambda^3_{\,\,\,1} & \Lambda^3_{\,\,\,2} & \Lambda^3_{\,\,\,3} \end{bmatrix} - \gamma^2 v^2 \begin{bmatrix} \Lambda^2_{\,\,\,2} & \Lambda^2_{\,\,\,3} \\ \Lambda^3_{\,\,\,2} & \Lambda^3_{\,\,\,3} \end{bmatrix} \end{align} to motivate choosing the rotation $R^{\alpha}_{\,\,\,\beta}$ so that the obvious relations $$\Lambda^1_{\,\,\,1} = \gamma$$ $$\Lambda^2_{\,\,\,2} = \Lambda^3_{\,\,\,3} = 1$$ $$\Lambda^1_{\,\,\,2} = \Lambda^1_{\,\,\,3} = ... = 0$$ make the above determinant relation an identity. Thus we work with $$ \Lambda^{\alpha}_{\,\,\,\beta} = \begin{bmatrix} \gamma & \gamma v & 0 & 0 \\ \gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$ Now, since the $3 \times 3$ spatial part of the matrix should reduce to $I$ when $\mathbf{v} = (v,0,0)$ is zero, we simply try to re-write it as in terms of the identity and a part that depends on $\mathbf{v}$ in a way that will easily generalize to arbitrary $\mathbf{v}$'s via \begin{align} \Lambda^{i}_{\,\,\,j} &= \begin{bmatrix} \gamma & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ &= I + (\begin{bmatrix} \gamma & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - I) \\ &= I + \begin{bmatrix} \gamma - 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \\ &= I + (\gamma - 1) \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \\ &= I + (\gamma - 1) (1,0,0) \otimes (1,0,0)^T \\ &= I + (\gamma - 1) \frac{1}{v^2} v(1,0,0) \otimes v (1,0,0)^T \\ &= I + (\gamma - 1) \frac{1}{\mathbf{v}^2} \mathbf{v} \otimes \mathbf{v}^T \\ &= \delta^i_{\,\,\,j} + (\gamma - 1) \frac{1}{\mathbf{v}^2} v^i v_j \end{align} This is Weinberg's (2.1.20), where I still have $\mathbf{v} = (v,0,0)$, but now the relation is a vector relation independent of the form of $\mathbf{v}$ so you can just set $\mathbf{v} = (v^1,v^2,v^3)$.

The choice $\mathbf{v} = (v,0,0)$ means $\Lambda^i_{\,\,\,0} = \gamma v^i$ reduces down to $\Lambda^i_{\,\,\,0} = \gamma v^i = (\gamma v,0,0)$ but it is easily generalized to a general $\mathbf{v}$ frame again by setting $\Lambda^i_{\,\,\,0} = \gamma v^i$ for general $\mathbf{v}$, which is (2.1.21).

The components $\Lambda^i_{\,\,\,j}$ cannot be uniquely determined. The best way you can motivate the form of these components are given in bolbteppa's answer. Perhaps this is the best you can do. However, this can still feel like cheating, especially when you generalize the results from $(v,0,0)$ to the case of $\mathbf{v}$. So, this answer will complement the previously cited answer in this generalizing. As we now know our convenient form the components $\Lambda^i_{\,\,\,j}$, we can do the following algebra,

\begin{equation} \begin{gathered} v_k\Lambda^k_{\,\,\,i}=\gamma v_i\\ v_k\Lambda^k_{\,\,\,i}=\left(\gamma-1 \right)v_i+v_k\delta^k_{\,\,\,\,i}\\ v_k\Lambda^k_{\,\,\,i}=\left(\gamma-1 \right)v_i\frac{v_kv^k}{\mathbf{v}^2}+v_k\delta^k_{\,\,\,\,i}\\ v_k\Lambda^k_{\,\,\,i}=v_k\left[\left(\gamma-1 \right)\frac{v_iv^k}{\mathbf{v}^2}+\delta^k_{\,\,\,\,i}\right]\\ v_k\left[\Lambda^k_{\,\,\,i}-\left(\gamma-1 \right)\frac{v_iv^k}{\mathbf{v}^2}+\delta^k_{\,\,\,\,i}\right]=0 \end{gathered} \end{equation} Now, as $v^k$ is arbitrary, we must have, \begin{align} \Lambda^k_{\,\,\,i}=\left(\gamma-1 \right)\frac{v_iv^k}{\mathbf{v}^2}+\delta^k_{\,\,\,\,i} \end{align}