Linear System given an equation with $Ax=b$ and where $A^tb=0$?

Question

Consider a system of linear equations $Ax=b$ where $A$ is an $n\times n$ matrix. Suppose that $b$ is a non-zero vector such that $A^t b=0$. Which is true about any such system?

I am given 5 choices for what this means: The system has infinitely many solutions, is inconsistent, is consistent, is over determined, or is in row echelon form.

Unfortunately, I have yet to completely rule out a single one. I don't think it can be over determined, because then wouldn't the equations be undefined based on the fact that if there were more equations than unknowns then $Ax$ would not be possible. I don't recall any theorems discussing whether the system is consistent or inconsistent using the transposition and b, which stems to the infinitely many solutions.

Answer 1

Suppose $Ax=b$ has a solution $x_0$. Then $A^tb=A^tAx_0=0$. This implies either $x_0^tA^tAx_0=||Ax_0||^2=0$ or $Ax_0=0$. So $b$ needs to be the zero vector.

Answer 2

I'll assume $A$ has real entries. A useful fact in this case is that $A$ and $A^TA$ have the same null space.

We are given that $b$ is a non-zero vector such that $A^T b=0$. Assume there exists a vector $x$ such that $Ax=b$. Multiply on the left by $A^T$ to see that $x \in N(A^T A)$. It follows that $x \in N(A)$. This implies that $b=0$, which is a contradiction.

Thus, the equation $Ax=b$ has no solutions.

Answer 3

Suppose $\|b\|_2=c$. Multiply the equation set with $b^t$ from the left

$$ 0=b^TAx=(A^tb)^tx=b^tb=c^2 $$ So there is no solution unless $c=0$ which is only possible if $b=0$ from the property of a norm.