I have been around this a couple times, and I haven't been able to fully understand this. Happens that I solved a space basis problem in my linear algebra course but I (and apparently anyone I have asked to) don't really get to understand what the result means. My question is, why the last row is a zero row? I understood something about a basis only spanning a three-dimensional subspace, but no more. Does that mean the fourth coordinate of each vector expressed as linear combination of the basis vector will always be zero such that the vector be in the subspace, or it doesn't work like that ?
So I'm going to provide a detailed explanation about the process I did (which is verified), and highlight the answer I got so someone could answer my question.
The problem is the following:
S = {v1, v2, v3, v4, v5} where: v1 = (1, 1, 0, -1) ; v2 = (0, 1, 2, 1) ; v3 = (1, 0, 1, -1) ; v4 = (1, 1, -6, -3) ; v5 = (-1, -5, 1, 0). Find a base for the subspace W = span S.
The way I learnt to find out if it was linearly independent and a spanning set was using Gauss-Jordan elimination. So here is my process:
$$ \begin{matrix} 1 & 0 & 1 & 1 & -1 & 0 & a \\ 1 & 1 & 0 & 1 & -5 & 0 & b \\ 0 & 2 & 1 & -6 & 1 & 0 & c \\ -1 & 1 & -1 & -3 & 0 & 0 & d \\ \end{matrix} $$
. . .
$$ \begin{matrix} 1 & 0 & 0 & 3 & -4 & 0 & (a + 2b - c)/3 \\ 0 & 1 & 0 & -2 & -1 & 0 & (-a + b + c)/3 \\ 0 & 0 & 1 & -2 & 3 & 0 & (2a - 2b + c)/3 \\ 0 & 0 & 0 & 0 & 0 & 0 & (4a - b - c + 3d)/3 \\ \end{matrix} $$
Here, I know the first three column vectors where the main entries are represent the basis for the supspace W = span S, it means, the vectors {v1, v2, v3}. So I tried to check if both of the conditions for a basis were fulfilled for the found basis.
(REASON WHY OF USING GAUSS-JORDAN ROW REDUCTION) I'm going to use row reduction to see if the only way to get the zero vector expressed as a linear combination of the basis is when all the constants are equal to 0. Also, to check if the basis is a generator set for every vector is the subspace. That's how I was teached to do it, but there's probably a faster and better way to do it.
$$ \begin{matrix} 1 & 0 & 1 & 0 & a \\ 1 & 1 & 0 & 0 & b \\ 0 & 2 & 1 & 0 & c \\ -1 & 1 & -1 & 0 & d \\ \end{matrix} $$
. . .
\begin{matrix} 1 & 0 & 0 & 0 & (a + 2b - c)/3 \\ 0 & 1 & 0 & 0 & (-a + b + c)/3 \\ 0 & 0 & 1 & 0 & (2a - 2b + c)/3 \\ 0 & 0 & 0 & 0 & (4a - b - c + 3d)/3 \\ \end{matrix}
As you can see, the last row is a zero row. I was expecting something like that due to the dimensions of the matrix. The thing I haven't yet understood is: i) why that row is a zero row? ii) Is it a basis or not? iii) In case it'd be a basis, does the fourth coordinate always has to be zero so a vector expressed as a linear combination of the basis vectors be contained in W? iv) does that mean that as W is 'embedded' in S, some vectors expressed as a linear combination of the basis vectors which fourth coordinate is not zero, the vector is in S but not in W?
I have a lot of questions about this, and I'm basically new in this topic, so I would appreciate so much someone who could easily explain this to me. Thank you.