also how did they know to create two separate vectors for s and t? Is it that a single variable goes for a single combination, meaning if x1 and x3 can are versions of one another then you can describe them both in terms of S, but x2 is not so it gets it's own variable t, and therefore it's own vector? But if t=0 then the second vector should be all zeros not 0,1,0 no?
Here the eigen space consists of vectors (x1 x2 x3) such that 1.x1 + 0.x2 + 1.x3 = 0. So, we find the values that x1, x2, x3 can have to satisfy the equation. We find that if x1= s, which is any scalar, then x3 must be -s and since the co-eff. of x2 is 0 it can have any value and still satisfy the equation and so we assign x2 the value t.
Now (x1 x2 x3) looks like (s t -s).
We need to express (s t -s) as a linear combination of linearly independent vectors.
(s t -s) = (s 0 -s) + (0 t 0) = s(1 0 -1) + t(0 1 0).
Here (1 0 -1) and (0 1 0) are linearly independent and from the above expression it it clear that any vector of the eigen space( which is of the form (s t -s)) can be expressed as a linear combination of these two vectors, in other words these two linearly independent vectors span the eigen space. So, (1 0 -1) and (0 1 0) form a basis for the eigen space.
Note: c(1 0 -1), where c is a scalar, is a scalar multiplication of (1 0 -1) which means any vector of the form c(1 0 -1) lies on the same straight line passing through (0 0 0) as (1 0 -1) does and they form a subspace of R3.
$\displaystyle \begin{bmatrix}1 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}x_1\\ x_2 \\ x_3\end{bmatrix}= \begin{bmatrix}x_1+ x_3 \\ 0 \\ 0 \end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix} $
reduces to the single equation $\displaystyle x_1+ x_3= 0$. That simplifies to $\displaystyle x_3= -x_1$ but there is NO condition on $\displaystyle x_2$. That means that $\displaystyle \begin{bmatrix}x_1 \\ x_2\\ -x_1\end{bmatrix}$ will satisfy the equation for any numbers, $\displaystyle x_1$ and $\displaystyle x_2$:
$\displaystyle \begin{bmatrix}1 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}x_1\\ x_2 \\ -x_1\end{bmatrix}= \begin{bmatrix}x_1- x_1 \\ 0 \\ 0 \end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix} $
no matter what $\displaystyle x_2$ is.
As before an eigen vector is of the form $\displaystyle \begin{bmatrix}x_1 \\ x_2 \\ -x_1\end{bmatrix}= \begin{bmatrix}x_1 \\ 0 \\ -x_1\end{bmatrix}+ \begin{bmatrix}0 \\ x_2\\ 0\end{bmatrix}= x_1\begin{bmatrix}1 \\ 0 \\ -1\end{bmatrix}+ x_2\begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}$
No, the second vector should not be "all zeros" because you are wrong about $\displaystyle x_2$ being 0. Since it is not required to satisfy any equation, it can be any number, not just 0.