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

The answer to this statement is false: No solution to $A\mathbf {x} =\mathbf {c}$ must exist (however, if a solution exists, it must be unique).

If $A\mathbf {x} =\mathbf {c}$ then $\mathbf {c}$ is a linear combination of the columns of $A$. However, the vector $\mathbf {c}$ is in $\mathbb {R} ^{5}$ 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 $A$ could properly represent any vector $\mathbf {c}$ and therefore we are not guaranteed that a solution exists to $A\mathbf {x} =\mathbf {c}$.

We can illustrate this with an example.

Let $A={\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0\end{bmatrix}}$, $\mathbf {b} ={\begin{bmatrix}1\\0\\0\\0\\0\end{bmatrix}}$ and $\mathbf {c} ={\begin{bmatrix}0\\0\\0\\0\\1\end{bmatrix}}$.

Then $A\mathbf {x} =\mathbf {b}$ has the unique solution $\mathbf {x} ={\begin{bmatrix}1\\0\\0\\0\end{bmatrix}}$, but $A\mathbf {x} =\mathbf {c}$ has no solution.

However, if a solution to $A\mathbf {x} =\mathbf {c}$ exists, then it must be unique: If $A\mathbf {x} =\mathbf {b}$ has a unique solution then it must have a trivial nullspace, i.e. there must not be a vector $\mathbf {y} \not =\mathbf {0}$ such that $A\mathbf {y} =\mathbf {0}$ as otherwise

$A(\mathbf {x} +\mathbf {y} )=A\mathbf {x} +A\mathbf {y} =\mathbf {b} +\mathbf {0} =\mathbf {b}$

and $\mathbf {x} +\mathbf {y}$ is a solution violating the uniqueness assumption of $A\mathbf {x} =\mathbf {b}$. So assume that $\mathbf {r}$ and $\mathbf {s}$ are both solutions of $A\mathbf {x} =\mathbf {c}$. We need to show that the solution is unique, that is, $\mathbf {r} =\mathbf {s}$. Subtracting the two equations we obtain $A\mathbf {r} -A\mathbf {s} =\mathbf {c} -\mathbf {c} =\mathbf {0}$ and thus $A(\mathbf {r} -\mathbf {s} )=\mathbf {0}$, that is, $\mathbf {r} -\mathbf {s}$ is in the nullspace of $A$. However, this nullspace only consists of the null vector, which implies that indeed $\mathbf {r} =\mathbf {s}$.