How do free variables become a constant that represents all span of a vector?

Question

Context: From this video(timestamped), making sense of null spaces:

I understand:

1. First statement from the left that we turned $x_2 = \begin{bmatrix}1 \\ 0\end{bmatrix} = s$ (yellow column).
2. second statement in green, $x_1 + x_2 = \begin{bmatrix}1 \\ 0\end{bmatrix} + \begin{bmatrix}1 \\ 0\end{bmatrix} = 0$
3. My Question: But I don't get how it became $\begin{bmatrix}x_1 \\ x_2\end{bmatrix}$, wouldn't that notation mean $\begin{bmatrix}\begin{bmatrix}1 \\ 0\end{bmatrix} \\ \begin{bmatrix}1 \\ 0\end{bmatrix}\end{bmatrix}$ and it to making sense at all?
4. I also know that we can sub $x_1 + x_2$ replacing $x_2 = s$, into $s = -x_1$, resulting in $s\cdot\begin{bmatrix}-1 \\ 1\end{bmatrix} $, which means all values of the span of $\begin{bmatrix}-1 \\ 1\end{bmatrix} $ (a line) is the answer(aka values that will become 0 or null)

I just don't understand point 3 and how it translates, how to write $\begin{bmatrix}x_1 \\ x_2\end{bmatrix}$ in numbers, please help me, and help pinpoint my knowledge gaps? LA is hard to teach one self...!

Answer 1

Let us consider real matrices only for simplicity. From the definition of null spaces, the null space of a matrix $A \in \mathbb{R}^{m \times n}$ is the set of vectors $\{x \in \mathbb{R}^n: Ax = 0\}$. Therefore, each vector $x$ in the null space of $A$ satisfies that $$\sum_{j \in [n]} A_{ij} * x_j = 0, \forall i \in [m].$$ For OP's example, since the second row consist of $0$'s only, we only need to care about the first row, which gives $$1 * x_1 + 1*x_2 = 0$$

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In particular, I don't think the first point OP wrote is correct, and that was why OP misunderstood the whole process.

Answer 2

Don't think of $x_2$ as being a vector $\begin{pmatrix}1\\0\end{pmatrix}$. The presenter doesn't say $x_2$ equals the second column. They only mean that the second column gives the coefficient of $x_2$ in each equation corresponding to that augmented matrix. $x_2$ is just a number in $\mathbb{R}$. You can refer to that number as $s$ if you like - that's what the presenter is trying to say. (They said $x_2=s$, but it would be clearer to say "Let $s:=x_2$", since that's the definition of $s$.)

The augmented matrix doesn't correspond to the equation $\begin{pmatrix}1\\0\end{pmatrix}+\begin{pmatrix}1\\0\end{pmatrix}=0$ - that doesn't make sense, the left-hand side is a vector in $\mathbb{R}^2$, and the right-hand side is a single number in $\mathbb{R}$. The augmented matrix also doesn't correspond to the equation $\begin{bmatrix}1\\0\end{bmatrix}+\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$ - that's just not true, it's the same as saying $\begin{bmatrix}2\\0\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$. Instead, each row of the matrix describes the coefficients of $x_1$ and $x_2$ in an equation. That is, in the first row, the first number (the first $1$) is the coefficient of $x_1$, and the second number (the second $1$) is the coefficient of $x_2$, and the number to the right of the bar (the $0$) is the constant on the other side. That is, the first row of the augmented matrix represents the equation $1x_1+1x_2=0$.

Given that $x_2$ is a real number, you know that there must be some $s \in \mathbb{R}$ satisfying $x_2=s$. You don't know what $s$ is - in fact, $s$ could be any real number - but you know that $x_2$ is equal to some number, and you can use the label $s$ to refer to that number. Then yes, you can turn $x_1+x_2=0$ into $x_1+s=0$.