Suppose are known and are unknown. Does the simultaneous system
have a unique solution
Answer & Explanation
vrtuljakc6
Beginner2022-07-12Added 16 answers
No, the solution is not unique, unless you have special configurations of the given bounds. Also, this is more of an exercise in basic real analysis than linaer algebra. If a solution exists at all, we have
and
and hence it must be true that
I'll ignore the degenerate cases where and/or , they don't bring anything new to the table (and fall under the special configurations I mentioned above), so I assume that the intervals whose intersection S is are proper intervals with positive length. is an interval, but it might be degenerate and just a point. Those are the 2 cases and . In the first case, it must be that and thus . The second case has similiar consequences, again we get . But again, this is a special configuration of the given bounds. If , then is a proper interval with . In this case, there are infinitely many and for each of them is a solution. That's because
and hence
and the same is true for . Moreover, because we chose from the open interval , we do not have equality in either side of the inequality above, so we get
That means is an inner point of the interval and so ther exists a such that
Now that means in the above solution we can replace with infintely many other pairs
for any Absolutely the same argument can be made for So unless you have initial bounds that trivially force your unknowns to be either bound of the interval, you will have infinitely many solutions, also with infinitely many of them such that and . To me it looks like what happens often when dealing with inequalities. You come to a point where you see something that would have nice consequences if it could be proven. But unfortunately, your system does generally not have a unique solution and it is not even close.