Meaning of linear independence with row vectors So far I have understood that a set of vectors $S = {v_1, v_2, . . . , v_k }$ in a vector space V is linearly independent
when the vector equation
$c_1v_1 + c_2v_2 + . . . + c_kv_k = 0$
has only the trivial solution$c_1 = 0, c_2 = 0, . . . , c_k = 0.$
An example in matrix form is:
$\begin{bmatrix}1 & 1 & 2 & 4 \\ 
0 & -1 & -5 & 2 \\ 
0 & 0 & -4 & 1 \\ 
0 & 0 & 0 & 6 \\ 
\end{bmatrix} 
\begin{bmatrix}
c_1\\ 
c_2 \\ 
c_3 \\ 
c_4 \\ 
\end{bmatrix} =
\begin{bmatrix}
0\\ 
0 \\ 
0 \\ 
0 \\ 
\end{bmatrix}
$
But a matrix of this form 
$\begin{bmatrix}1 & 1 & 2 & 4 \\ 
0 & -1 & -5 & 2 \\ 
0 & 0 & -4 & 1 \\ 
0 & 0 & 0 & 0 \\ 
\end{bmatrix} 
\begin{bmatrix}
c_1\\ 
c_2 \\ 
c_3 \\ 
c_4 \\ 
\end{bmatrix} =
\begin{bmatrix}
0\\ 
0 \\ 
0 \\ 
0 \\ 
\end{bmatrix}
$
is linearly dependent because it has more than a trivial solution
However, I am confused about row vectors, specifically the idea that to get a basis for a subspace using row vectors we must put the matrix in reduced row echelon form to find the linearly independent vectors. For example here the accepted answer gives an example of finding a basis with row vectors using this
$\begin{bmatrix}1 & 1 & 2 & 4 \\ 
2 & -1 & -5 & 2 \\ 
1 & -1 & -4 & 0 \\ 
2 & 1 & 1 & 6 \\ 
\end{bmatrix} \Rightarrow \begin{bmatrix}1 & 1 & 2 & 4 \\ 
0 & -3 & -9 & -6 \\ 
0 & -2 & -6 & -4 \\ 
0 & -1 & -3 & -2 \\ 
\end{bmatrix} \Rightarrow \begin{bmatrix}1 & 1 & 2 & 4 \\ 
0 & -3 & -9 & -6 \\ 
0 & 0 & 0 & 0 \\ 
0 & 0 & 0 & 0 \\ 
\end{bmatrix}$
and then goes on to say that "Only two of the four original vectors were linearly independent." In what respect are these two vectors linearly independent? This looks exactly like the second example that I gave in which the vectors were dependent because they had more than a trivial solution? Does linear independence with regard to row vectors mean something else? Or does this also only have a trivial solution, and if so, how?

Yes, it is "exactly like the second example" you gave.  In that example, you were showing that the original four vectors were dependent.  But there will always be some subset of those vectors that is independent.  (That subset might consist of a single vector.  A set containing just one vector is always "independent".)  In this last problem you are looking for that independent subset.
Your "four vectors" are the rows of the matrix (1, 1, 2, 4), (2, -1, -5, 2), (1, -1, -4, 0), and (2, 1, 1, 6).  To determine whether they are "independent" or "dependent", you look for numbers, a, b, c, d, such that a(1, 1, 2, 4)+ b(2, -1, -5, 2)+ c(1, -1, -4, 0)+ d(2, 1, 1, 6)= (a+ 2b+ c+ 2d, a- b- c+ d, 2a- 5b- 4c+ d, 4a+ 2b+ 6d)= (0, 0, 0, 0).  So we have the four equations, a+ 2b+ c+ 2d= 0, a- b- c+ d= 0, 2a- 5b- 4c+ d= 0, and 4a+ 2b+ 6d= 0.  Adding the first two equations eliminates c: 2a+ b+ 3d= 0.  But notice that the fourth equation, 4a+ 2b+ 6d= 0, is just two times that so is not really a new equation. If we add the third equation to 4 times the first equation we also eliminate c: 6a+ 3b+ 9d= 0. We can divide by 3 to get 2a+ b+ 3d= 0.
We have two equations, 3a+ b+ 3d= 0 and 2a+ b+ 3d= 0.  Subtracting the second of those from the first gives a= 0.  But then we have b+ 3d= 0 so b= -3d but d can be anything.  Putting a= 0, b= -3d into a+ 2b+ c+ 2d= 0 we have -6d+ c+ 2d= c- 4d= 0 so c= 4d.  That means that a(1, 1, 2, 4)+ b(2, -1, -5, 2)+ c(1, -1, -4, 0)+ d(2, 1, 1, 6)= (0, 0, 0, 0) can be written -3d(2, -1, -5, 2)+ 4d(1, -1, -4, 0)+ d(2, 1, 1, 1, 0)= (0, 0, 0, 0).  Then d(2, 1, 1, 1 0)= 3d(2, -1, -5, 2)- 4d(1, -1, -4, 0) or, dividing by d, (2, 1, 1, 0)= 3(2, -1, -5, 2)- 4(1, -1, -4, 0).  So those 4 vectors can actually be written in terms of these last two. These two vectors form an independent subset of the original four vectors.  The space they span is actually two dimensional, not four dimensional.  Using "row reduction" with a matrix is just a more convenient way of doing that.
