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In cases where linear dependency or independency of some number of vectors is to be found out, one of the methods is to form a matrix of the vectors of dimensions m×n where the column represents the coordinates of a specific vector and the row represents the coordinates of all the vectors along a specific dimension .This is then multiplied with a coefficient matrix and then equated to 0. Solving such problems involve converting the matrix into its row echelon form via row operations. My question here is,why exactly can we perform row operations on the matrix?It is not like we are multiplying one vector with some scalar and add (or subtracting) which is usually the procedure without using matrices.It is like multiplying coordinates of all vectors along specific dimension with some scalar and performing operations on coordinates of all the vectors along other dimension.This is what is shown in my textbook.why does this work?

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The operations are equivalent to multiplying by elementary matrices which are invertible. Therefore, if E is an elementary matrix, $$ A\cdot\vec{v} = \vec{b} \iff E \cdot A\cdot\vec{v} = E \cdot \vec{b} $$

because we can go back by multiplying by $E^{-1}$, and thus the original system is equivalent to the last one which is multiplied by $E$. Repeating this process by multiplying by different elementary matrices leaves us with an upper-triangular matrix which is equivalent to the original.

Esentially, when checking if a vector set is linearly independent, you're searching for non-trivial linear combinations of these vectors that equal zero, which we can compactly express as "does $A\cdot v = \vec{0}$ has non-zero solutions?" where $A$ is a matrix that has the vectors as columns.

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  • $\begingroup$ While that is correct.One thing that still doesn't get cleared is that how/why can we do this in case of the vector matrix. Because to solve the problem, we multiply the whole vector with some scalar which involves multiplying all the coordinates of a vector with a scalar . Row operations quite do the opposite.They change individual coordinates. To solve this,it should've probably been that each row represents an individual vectors' coordinates. Then the operations hold true. $\endgroup$ – nalostta Dec 5 '16 at 14:22
  • $\begingroup$ Hmm, I understand what you're saying, though the traditional method is by row-operations. Nevertheless, you can check by hand yourself that if $v_1,... v_k$ are linearly independent, so are $v1, ... \alpha \cdot v_k$. $\endgroup$ – guidoar Dec 5 '16 at 14:24
  • $\begingroup$ Can u pls type out the last sentence again...I can't make out it just shows $v_1....and a bunch of some other characters $\endgroup$ – nalostta Dec 5 '16 at 16:51
  • $\begingroup$ The last thing I said was that, if you have a set of vectors, multiplying one of them by a scalar will not change their linear dependence (or lack of). This is easy to prove using the definition of linear dependence, see if you can prove it yourself. $\endgroup$ – guidoar Dec 5 '16 at 18:21

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