I have following problem:
Let $A,B \in F^{n\times n}$, $B$ is regular. Which on of the following is true?
1.) If $(A|b)$ has at least one solution, then $(BA|b)$ has at least one solution.
2.) If $(A|b)$ has at least one solution, then $(AB|b)$ has at least one solution.
3.) If $(BA|b)$ has at least one solution, then $(A|b)$ has at least one solution.
4.) If $(AB|b)$ has at least one solution, then $(A|b)$ has at least one solution.
($(A|b)$ is a set of linear equations, not sure if that's clear)
If my logic is correct, 1 and 3 are true, and 2 and 4 are false.
Because regular matrix cannot change number of solutions, the multiplication should have no effect in this implication. However this is not true the other way around, if we multiply with not regular matrix, there is a possibility, that the number of solutions may change.
However the answer feels way too easy, so i am not sure if i am missing something obvious here.
Thanks for your time!
So, apparently this is a little bit too late, also they do not tend to change the homeworks on our university too much across the classes. Maybe it will be of help to someone else.
Up to the topic.
1) is not true, as I can think of a certain pick of n=2 and two matrices and a vector of right side, which would be A=((1,1),(1,1)) in row form (obviously singular) B=((1,0),(-1,1)) in row form. b = (1,1)^T. Matrix B just so happens to be the elementary row operation matrix which adds -1*1st row of matrix A to the 2nd row of matrix A.
Result is BA = ((1,0),(-1,1))((1,1),(1,1)) = ((1,1),(0,0)). Now take look at (B*A|b). It is now a system of linear equations in form x + y = 1 0 + 0 = 1 Which obviously does not have solution at all (as a direct result of Frobenius' lemma).
2) This is true, as multiplying just any matrix A with regular matrix B from RIGHT side does not change the COLUMN space of A (aka. the set of vectors, which can be generated using columns of matrix A). The proof is rather long, but to put things simple
it is very alike for multiplying just any matrix A with regular B from LEFT side. That does not change the ROW space of matrix A (which goes precisely from the definition). We commonly do that for solving lineary (in)dependence of matrices and we call it elementary row operations (in fact any elementery row operation can be written as multiplying the given system by regular matrix from left).
We just concluded, that multiplying with regular matrix B from right does not change the COLUMN space. Therefore Im(AB)=Im(A). Thus vector b^T has to be element of both Im(AB) and Im(A).
3) Just like in 1) I present a counterexample which follows me thinking of a n=2 and matrix B*A=((1,1),(0,0)) a right side vector b^T. I now want to find matrix A, which would be in form ((1,1),(0,0)). What is matrix B?
We could solve this now, but the solution is the same as in 1), as I used the exact same matrices. B=((1,0),(-1,1)).
Check the starting system in this case of scenario (it has solution) and check (A|b), which is yet again in form of x + y = 1 0 + 0 = 1 We conclude through Frobenius' lemma, that this system has no solution.
4) We will use the same trick, but let us rename AB = C. Now we know, for a fact, that matrix B is invertible (it is regular). We find B^(-1), which is yet again regular (having inverse matrix (B^-1)^-1 = B).
Now we suppose that (AB|b)=(C|b) has a solution. Using the same logic as in case 2), we could just say that Im(AB)=Im(A) and be done with it. Or we could multiply our matrix C from the left with regular matrix B^(-1), we already know, that this does not change the column space of C. Therefore (using 2)) Im(C)=Im(C*B^(-1)).
Let us remember that C=AB and look at the result. Im(ABB^(-1))=Im(A)=Im(AB). Concusion is, that Im(A)=Im(AB), thus the statement holds true.
PS: For any students reading this: I encourage you to think through why you can identify elementary row operations with multiplying by regular matrices from right. Also try to look up the proof that multiplying matrix A with regular B from left does not change the column space.