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Given $n$ vectors, we want to determine if those vectors are linearly independent.

One way doing it is writing those vectors as columns of a matrix and row-reduce it.
The vectors are linearly independent iff there's only the trivial solution for the homogeneous system.

Now for my understanding, One can also write those vectors as rows of a matrix and row-reduce the matrix: those vectors are linearly independent iff there's no zeros-vector.

Is there a "connection" between those two methods (maybe in terms of "row/cols space")?
Thanks.

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    $\begingroup$ "Now for my understanding, One can also write those vectors as rows of a matrix and row-reduce the matrix: those vectors are linearly independent iff there's no zeros-vector." This is only true with the additional hypothesis that the vectors are $n$-dimensional. Counterexamples are easy to come by. $\endgroup$ – Git Gud Jun 6 '14 at 10:26
  • $\begingroup$ You should read about a related topic: en.wikipedia.org/wiki/Rank_%28linear_algebra%29 $\endgroup$ – M. Vinay Jun 6 '14 at 10:34
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Yes, indeed, there is a relationship between the two.

The rank of a matrix $A$ is the number of linearly independent columns of $A$ (column rank), which is also the number of linearly independent rows of $A$ (row rank).

Equivalently, the column rank of $A$ is the dimension of the column space of A, while the row rank of $A$ is the dimension of the row space of $A.$

For every matrix, the column rank is equal to the row rank.

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