Why doesn't transposing matter? I was helping a friend with linear algebra,
particularly I was teaching how to check if a collection of vectors $v_1, ..., v_k \in \mathbb{R}^n$ are linearly independent (assuming the vectors are given vertically). Now the techniques I gave him was to apply reduced row echelon form to the matrix
$$ \begin{bmatrix} v_1^T \\ v_2^T \\ \vdots \\ v_k^T  \end{bmatrix} $$ 
And if one doesn't discover any "zero" rows then the vectors are indeed linearly independent. 
So as an example, verifying that:
$$\begin{bmatrix}  1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix}  1 \\ 0 \\ 0 \end{bmatrix} $$
Are linearly independent occurs by applying the reduced row echelon form to
$$ \begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 0  \end{bmatrix} $$ 
yielding
$$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0  \end{bmatrix} $$
So we conclude that there are no dependent vectors.
Now my friend suggests that running the same routine, but without transposing the vectors can also let us conclude about dependence.
His claim:
The number of non zero rows in:
$$ \begin{bmatrix} v_1 & v_2 & ... & v_k\end{bmatrix} $$ 
Corresponds to the number of dependent vectors. I disagreed with a counter example, to which he responded with an argument involving "counting pivots" that I couldn't make sense of.
Has anyone here seen something like that? Would they be able to enlighten me on how to test independence/dependence without taking transposes?
 A: The column- and row rank of a matrix coincide, so it does not matter whether you transpose the matrix or not.
A: *

*Elementary row operations alter the linear dependences among the rows, but they do not alter the row space. Clearly the new rows introduced by any row operations are in the row space of the old rows.  And vice-versa, since row operations can be inverted by other row operations.  Since they do not alter the row space, they do not alter the dimension of the row space.

*Elementary row operations alter the columns space, but they do not alter the linear dependences among the columns. Consider $$ \left[ \begin{array}{rrr} m_{1,1} & \cdots & m_{1,c} \\ \vdots\ \ \ \ & & \vdots\ \ \ \ \\ m_{r,1} & \cdots & m_{r,c} \end{array} \right] \left[ \begin{array}{c} x_1 \\  \vdots \\ x_c \end{array} \right] = \left[ \begin{array}{c} 0 \\ \vdots \\ 0  \end{array} \right] $$ and suppose the $x$s are not all $0$.  Then the vector of $x$s is a linear dependence among the columns. If that vector is a solution to the system before the application of a row operation, then it still is afterwards.  Since the linear dependences therefore do not change, the dimension of the column space does not change.
Once the matrix is fully reduced, it is easy to see that the dimensions of the row and column spaces are equal.  Since they are equal after the row operations, and the row operations don't change them, they are equal before the row operations. Conclusion: In every matrix, the row and column spaces have equal dimensions. 
