$A$ is a real $n\times n$ matrix. If $Ax=0$ has infinitely many solutions, does $A^Tx=b$ and $b\not=0$ also have infinitely many solutions? I can't think of a counterexample.
Thank you for your help everyone.
2026-03-27 14:28:40.1774621720
If $Ax=0$ has infinitely many solutions, does $A^Tx=b$ also have infinitely many solutions?
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If $Ax=0$ has infinitely many solution this means that A is singular that is $rank(A)<n$ and thus also $rank(A^T)<n$.
Therefore $A^Tx=b\neq0$ has infinitely many solution or no solution.