Prove that $AX = B$ Is Consistent

Question

Suppose that $A$ is $n \times m$ and $B$ is $n \times p$. We are given that: $$c^tA = 0 \ \ \Leftrightarrow \ \ c^tB = 0$$ The text says that this relation implies that $AX = B$ is consistent since each column of $B$ is a linear combination of the columns of $A$. How is that so? To me, $c^tA$ looks more like a linear combination of the rows of $A$.

Answer 1

By fundamental theorem of linear algebra,

$$N(A^T)=R(A)^\perp$$

that is the nullspace of $A^T$ is equal to the orthogonal complement of $R(A)$, the range of $A$.

The information given in the question tells us that

$$N(A^T)=N(B^T)$$

Hence

$$R(A)^\perp=R(B)^\perp$$

Since $R(A)$ and $R(B)$ are finite dimensional,

$$R(A)=R(B).$$

Hence $AX=B$ is consistent.