Say I have a matrix $A \in \mathbb{R}^{m \times n}$ where $m \leq n$ (in other words the matrix is either square or "wide"). Let $A$ have full rank (meaning $rank(A) = \min\{m,n\} = m$). If this matrix is square then clearly by the uniqueness of the inverse, $Ax=b$ has one solution for a given vector $b.$ If the matrix is "wide" ($m < n$) then obviously $Ax=b$ can have infinitely many solutions. My question is, can this "wide" system ever have either exactly one solution or no solution?
[Math] Number of solutions for full rank, wide matrix
linear algebra
Best Answer
No. We can always choose $n-m$ variables arbitarily and get a solution dependent on the choice.