elementary row operations effects

Question

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?

Answer 1

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."