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

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

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

We can illustrate this with an example.

Let $A = [\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{matrix}]$ , $𝐛 = [\begin{matrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix}]$ and $𝐜 = [\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{matrix}]$ .

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

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

A (𝐱 + 𝐲) = A 𝐱 + A 𝐲 = 𝐛 + 𝟎 = 𝐛

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