Rank nullity theorem proof So I'm watching a proof about the rank nullity theorem and it says this:

Theorem. Let Ax=0 be a homogeneous system with n variables.
Then the dimension of the nullspace is n-r(A).
Proof: Denote k=n-r(A). k is the number of "degrees of freedom", so there are k $x_{i}$'s we can choose. Assume for simplicity that they are $x_{1}, x_{2} ... x_{k} $.

He then says this:

A solution looks like this:
$\begin{bmatrix} x_{1} \cr x_{2} \cr ...\cr x_{k}\cr a_{1}x_{1}+a_{2}x_{2} +... a_{k}x_{k}\cr b_{1}x_{1}+b_{2}x_{2} +... b_{k}x_{k}\cr ...\cr  c_{1}x_{1}+c_{2}x_{2} +... c_{k}x_{k}   \end{bmatrix}$

Now I'm honestly really confused because, well, the way he said it, that all the rows after the $x_{k}$th row are just linear combinations of the previous rows. But then I have a question, is this just a general solution to the equation $Ax = 0$, right? Why do you need linear combinations here? Like there must be some thing I'm not understanding, because all these rows are just going to be numbers when I solve for the vector, right? So would I not be able to represent it with just a scalar multiple of x1 for every single row? Like I really do not understand this. Could anyone help?
 A: (A previous version of the answer, which misinterpreted the context, can be found in the edit history)

The statement you are confused about is important on its own and can be discussed separately from the rank-nullity theorem.
Proposition. A solution of a homogeneous linear system $\mathbf{Ax}=\mathbf{0}$ is of the form
$$\mathbf{x} = \left[\begin{array}{c}
x_{1}\\
\vdots\\
x_{k}\\
\alpha_{1}x_{1}+\dots+\alpha_{k}x_{k}\\
\vdots\\
\omega_{1}x_{1}+\dots+\omega_{k}x_{k}
\end{array}\right],$$
where $x_i$ are arbitrary numbers and all $\alpha_{i}, \dots,\omega_i$ are defined by $\mathbf{A}$.
This statement is probably proved or explained in Lecture 17. In a nut shell, by applying certain invertible row operations, you can transform any matrix $\mathbf{A}$ to the reduced-row-echelon form (RREF, also called "canonical" in these lectures). In this RREF, the number of "leading 1's" equals the number of non-zero rows equals the number of basic columns. Either of these numbers can be taken as a definition of the rank of a matrix, $r(\mathbf{A})$. The complete set of solutions to a linear system (aka the general solution) can be easily described by looking at the reduced-row-echelon form of the matrix (and the transformed right-hand-side, which is will be all zeros for a homogeneous system): some solution components are determined by the RREF and the right-hand side (there are $r(\mathbf{A})$ of them), the rest can be chosen arbitrarily (there are $n - r(\mathbf{A})$ of them). By reordering the vector components (together with the columns of the matrix), the general solution can be put in the above form, where $x_i, i=1\dots k$ are arbitrary.
This proposition is the most important to understand. It involves many non-trivial aspects, like the RREF, its existence, uniqueness, and why any solution is of this form. Afaic, some of them are not rigorously proved in these lectures, but even when taken on faith, they still deserve consideration.
Once you thoroughly understand the proposition, the rank-nullity theorem follows almost trivially. One may say that the proposition is much more involved and complex, whereas the rank-nullity theorem provides a high-level perspective on the image (aka the range) and null-space (aka the kernel) of a matrix as a linear operator.
