Let be the set of all b's which make solvable.
Consider that for an arbitrary and we have and .
If we take the sum .
In the same way, we can have, for , .
I hope this is what you're looking for.
Hi, I've been trying to figure out this problem, but I'm not quite sure how to go about it.
For any matrix A(m,n) show that the set of right hand sides b in R^m for which Ax = b is solvable is a subspace of R^m
Attempt at a solution:
I know how to prove the solvability of a specific linear system, but how can I show that is true for any matrix? (Assuming I need to)
Any help/advice will be much appreciated!
You might give some consideration to the column space of A, which we (should) know is a subspace of R^m.
Can such a b fail to reside in the column space of A?
Is there a vector in the column space of A, say b, where we find that there is no solution set in R^n for the system Ax=b?