The constraints on the full model can be written as $$\mathbf{L} 𝛃 = \mathbf{c} $$ Using Lagrange multiplier, we can get the estimate for null hypothesis $\widehat{𝛃}_0$ by minimizing $$\mathcal{L}\equiv\left(\mathbf{y}-\mathbf{X}𝛃\right)^{T}\left(\mathbf{y}-\mathbf{X}𝛃\right)+\left(\mathbf{L}𝛃-\mathbf{c}\right)^{T}𝛌,$$ where $𝛌$ is a column vector of multipliers. The minimization leads to $$ -2\mathbf{X}^{T}\left(\mathbf{y}-\mathbf{X}𝛃_{0}\right)+\mathbf{L}^{T}𝛌_{0}=0 $$ and $$ \mathbf{L}𝛃_{0}=\mathbf{c}. $$ These equations give \begin{align*} & \mathbf{L}^{T}𝛌_{0}=2\mathbf{X}^{T}\left(\mathbf{y}-\mathbf{X}𝛃_{0}\right)=2\mathbf{X}^{T}\mathbf{y}-2\mathbf{X}^{T}\mathbf{X}𝛃_{0}\\ \Rightarrow\, & \left(\mathbf{X}^{T}\mathbf{X}\right)^{-1}\mathbf{L}^{T}𝛌_{0}=2\widehat{𝛃}-2𝛃_{0}\\ \Rightarrow\, & \mathbf{L}\left(\mathbf{X}^{T}\mathbf{X}\right)^{-1}\mathbf{L}^{T}𝛌_{0}=2\mathbf{L}\widehat{𝛃}-2\mathbf{c} \end{align*} or $$ 𝛌_{0}=2\left(\mathbf{L}\left(\mathbf{X}^{T}\mathbf{X}\right)^{-1}\mathbf{L}^{T}\right)^{-1}\left(\mathbf{L}\widehat{𝛃}-\mathbf{c}\right), $$ as well as $$ 𝛃_{0}=\widehat{𝛃}-\left(\mathbf{X}^{T}\mathbf{X}\right)^{-1}\mathbf{L}^{T}𝛌_{0}/2, $$ or \begin{align*} \widehat{𝛃}-𝛃_{0} & = \left(\mathbf{X}^{T}\mathbf{X}\right)^{-1}\mathbf{L}^{T}𝛌_{0}/2\\ & = \left(\mathbf{X}^{T}\mathbf{X}\right)^{-1}\mathbf{L}^{T}\left(\mathbf{L}\left(\mathbf{X}^{T}\mathbf{X}\right)^{-1}\mathbf{L}^{T}\right)^{-1}\left(\mathbf{L}\widehat{𝛃}-\mathbf{c}\right). \end{align*}