This is a proof from a textbook

enter image description here What I don't undersdand is, clearly the cofficients for r_i is not equal, unless a_j is 0 (k has to be non-zero), but we want a_j to be any number, don't we? so a_i is not equal to a_i + k*a_j, which means vector whose a_j is non-zero in the original spanning sets cannot be a linear combination of the new spanning sets, then how can we say row operations do not change the row space of a matrix?

  • $\begingroup$ Given any linear combination of $r_1,\dots,(r_j+kr_i),\dots,r_n$, this can be written as a linear combination of $r_1,\dots,r_n$ by moving the term $a_j k r_i$ and grouping it with the other $r_i$ term. That's all that matters, the fact that $a_i+ka_j$ isn't the same as $a_i$ doesn't matter. $\endgroup$ – Ian Apr 24 '16 at 2:26

This is an example of a simple thing made complicated by formalism. Consider the following geometric analogy. Imagine the physical 3-dimensional space and two linearly independent vectors. Now imagine the plane that the two vectors span. Three questions:

  1. Does the plane the vectors span change when you switch them?
  2. Does the plane the vectors span change when you multiply one of them by a number (other than zero)?
  3. Does the plane the vectors span change when you add a multiple of one vector to the other?

With a little bit of a geometric imagination the answer is clearly "no" to each of the questions, more or less because in each case you end up with two vectors which may be different from the original vectors, but they still lie in the same plane and are still linearly independent.

Now the row space of a matrix consists of vectors of an entirely different nature, but logically there is no difference.


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