$Ax=b$ has two possible difficulties.
(1)Dependent rows
$Ax=b$ may have no solution when $b$ is outside the column space.
Instead of $Ax=b$, we solve $A^{T}A\hat{x}=A^{T}b$.
(2)Dependent columns
In this case, the solution of $A^{T}A\hat{x}=A^{T}b$ may not be unique.
The optimal solution of $Ax=b$ is the minimum length of $A^{T}A\hat{x}=A^{T}b$.
I can't understand the second one.
If $A$ has dependent columns, why the solution of $A^{T}A\hat{x}=A^{T}b$ may not be unique?