Science:Math Exam Resources/Courses/MATH221/April 2010/Question 06/Solution 1

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The answer to this statement is false: No solution to must exist (however, if a solution exists, it must be unique).

If then is a linear combination of the columns of . However, the vector is in and this space requires 5 vectors to form a basis. We can therefore not guarantee that the 4 vectors we have from the columns of could properly represent any vector and therefore we are not guaranteed that a solution exists to .

We can illustrate this with an example.

Let , and .

Then has the unique solution , but has no solution.

However, if a solution to exists, then it must be unique: If has a unique solution then it must have a trivial nullspace, i.e. there must not be a vector such that as otherwise

and is a solution violating the uniqueness assumption of . So assume that and are both solutions of . We need to show that the solution is unique, that is, . Subtracting the two equations we obtain and thus , that is, is in the nullspace of . However, this nullspace only consists of the null vector, which implies that indeed .