I am trying to look at the affect that elementary row operations have on a given matrix (with the notion that elementary row operations are linear combinations of the rows ):

  • $Rspan(A)$ is not changed
  • $Cspan(A)$ is changed $\rightarrow$ $Im(T_A)$ is changed
  • $Rank(A)$ is not changed
  • $Ker(T_A)$ is not changed

and in particular if elementary row operations do effect the $Cspan$ why doesn't it change the solution set? (Update: I was not doing elementary row operations on the column $b$, that is why the solution set is not changed)

Is that right?


Everything you've said is correct.

You ask "why, if it affects the column span, does it not change the solution set?"

Well, if you write a row op as left-multiplication by some elementary matrix $R$, then the solutions to $$ (RA)x = b $$ are indeed quite different from those for $$ Ax = b, $$ so row operations do change the solution set. The good news is that a slight variation of this is still useful. If you want to solve $$ Ax = b $$ you can instead solve $$ (RA)x = Rb $$ i.e., if you do the same row op to $A$ and to the target vector $b$, the solution set remains unchanged. If you do lots of row ops to make $A$ become diagonal, or upper triangular, the system may then be easy to solve. (Indeed, this is sometimes called something like "augmented Gaussian elimination", because you stick the column vector b on the right hand side of the matrix A, and perform row ops on the whole mess.)

In the special case $b = 0$, the row ops have no effect on $b$, and hence solving $$ Ax = 0 $$ and $$ (RA)x = 0 $$ give the same results, which is what you've observed in writing "$Ker(T_A)$ is not changed."

  • $\begingroup$ Thank, I just noticed that i was not doing row operations on column b $\endgroup$ – gbox Feb 13 '15 at 13:46

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