I'm trying to understand the approach described in these questions:
How to apply non zero Dirichlet boundary condition in finite elements?
While I understand what's happening, I don't understand why the approach is justified.
My understanding of what's happening:
Throughout this post, I will use the notation of $a_{0,..,n}$ denoting the $n \times 1$ matrix of elements $(a_0, ..., a_n)$.
Here is what I see as the a description of the actual solution - suppose we have the system $Ax = 0$ where $A$ is a $n \times n$ matrix, and suppose $x_0 =1 $ by the boundary condition, and all the other $x_i$ are free. Denote $A$ as \begin{bmatrix} c_0 & c_{1,...,n}^T\\ c_{1,...,n} & K \end{bmatrix}
Then $A(x_0e_0 + \Sigma_{i>0}x_i e_i) = A(x_0e_0) + A(\Sigma_{i>0}x_i e_i) = 0$ and so $A(\Sigma_{i>0}x_i e_i) = -A(x_0e_0) = -A(e_0)$, in other words: \begin{equation*} \begin{bmatrix} c_{1,...,n}^T\\ K \end{bmatrix} * (x_{1,...,n}) = -(c_{0,...,n}) = - \begin{bmatrix} c_0\\ c_{1,...,n} \end{bmatrix} \end{equation*} and so we essentially end up solving a system $\tilde{A}\tilde{x} = \tilde{b}$ where $\tilde{A}$ is an $n \times (n-1)$ matrix.
However, the approaches in the linked questions reach a different equation: \begin{equation*} K * (x_{1,...,n}) = -(c_{1,...,n}) \end{equation*}
My question:
Essentially, in the linked approaches, the first row is zeroed out, and I don't quite understand how that makes sense. You can zero the first column, sure, as long as you move it to the right hand side, but I'm not sure how it makes sense to zero out the whole row too - if there are multiple solutions to $K * (x_{1,...,n}) = -(c_{1,...,n})$, we need to find the one that ends up giving $-(c_{0,...,n})$.
How is this step justified?