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.
Problem:
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're given A is mxn.
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?