Is there a way to interpret the condition of unfeasibility of a linear system?

Question

A linear system $A x = b$ is not solvable if and only if there exists a vector $q \in \mathbb R^n $ such that $$ q^T A = 0^T \text { and } q^T b \neq 0 $$

Is there a geometrical translation to this statement?

Answer 1

Rephrasing a bit:

Let $\mathrm A \in \mathbb R^{m \times n}$ and $\mathrm b \in \mathbb R^m$. Show that the linear system $\rm A x = b$ is not solvable if and only if there exists a vector $\mathrm q \in \mathbb R^m$ such that $\mathrm q^\top \mathrm A = \mathrm 0_n^\top$ and $\mathrm q^\top \mathrm b \neq 0$.

A sketch of the proof of the $\impliedby$ direction:

• From $\mathrm q^\top \mathrm b \neq 0$ we conclude that $\mathrm q \neq 0_m$.

• From $\mathrm q^\top \mathrm A = \mathrm 0_n^\top$ we conclude that $\rm q$ is orthogonal to every column of $\rm A$, i.e., vector $\rm q$ is orthogonal to the column space of $\rm A$, i.e., vector $\mathrm q$ is in the left null space of $\rm A$. Since we already concluded $\mathrm q \neq 0_m$, we can now conclude that the column space of $\rm A$ is not $\mathbb R^m$ (i.e., the left null space is not trivial).

• From $\mathrm q^\top \mathrm b \neq 0$ we conclude that vector $\rm b$ is not in the column space of $\rm A$. In other words, there is no linear combination of the columns of $\rm A$ that produces $\rm b$. Therefore, $\rm A x = b$ does not have a solution. It has a least-squares solution, however.

The $\implies$ direction is left as an exercise for the reader.