Is the calculated solution, if it exists, unique?

Question

I am looking the following:

We have the system of equations \begin{equation*}Ax=\begin{pmatrix}1 & 3 & -4 \\ 3 & 9 & -2 \\ 4 & 12 & -6 \\ 2 & 6 & 2 \\ 0 & 1 & 0\end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix}=\begin{pmatrix}1 \\ 2 \\ 3 \\ 1\\ 1\end{pmatrix}=b\end{equation*} Calculate the system $Ax=b$ with the least squares method. Is the calculated solution, if it exists (show that!), unique?

$$$$

I have done the following:

To solve the system by the least squares method we solve the system $A^TA\overline{x} = A^Tb$. Using the Gauss-Algorithm we get the solution $\overline{x}=\begin{pmatrix}-2.4 \\ 1 \\ -0.1\end{pmatrix}$.

About the second part of the question, what exactly is meant? Since we have calculated a solution, how can we show if a solution exists? Or have I understood the question wrong?

Answer 1

What you might find useful, is to try finding a non-trivial $y \ne 0$ solution of $Ay = 0$. If such $y$ exists, then for any constant $C$ the solution $x$ of the equation $Ax = b$ is not unique, since

$$ A(x + Cy) = Ax + C Ay = Ax + C \cdot 0 = Ax = b $$

which presumes non-uniqueness.

Otherwise, if there is no such $y: Ay = 0$, other then $y = 0$, then your solution is unique.

Answer 2

[Collecting the discussion in comments into an answer.]

You’ve already proven existence of a solution by producing one. As for uniqueness, inspect the row-reduced matrix that you computed to find this solution: it has no rows without a pivot, so the solution is unique—there are no free variables in the solution.

The original system may or may not be solvable for other values of $\mathbf b$. For there to be a solution at all, we must have $\mathbf b\in\operatorname{Col}(A)$. You might be able to determine this by inspection for some values of $\mathbf b$, but in general testing this condition amounts to solving the system. On the other hand, the least-squares system $A^TA\mathbf x=A^T\mathbf b$ will always have a solution for this $A$: its columns are linearly independent, so $A^TA$ is a nonsingular square matrix, hence $(A^TA)^{-1}$ exists.