On the linear system $A x = 0$ when $A$ is rank-one

Question

From Strang's textbook:

---

---

For rank-one matrices, we have that $u(v^Tx)=0 \implies v^Tx=0$. Why does this follow?

Answer 1

Here's my attempt to make some sense of the question. Note that I am blaming Strang, not Curious Penguin. It is not Curious Penguin's fault that she/he is struggling to make sense out of Strang's vague and obscure remarks. It appears that $u,v \text{ and }x$ are $n-$ place column-vectors (We can assume $n>1.$) over some field $F.$ If $u$ or $v$ is the zero vector, we have a complete triviality, so let's assume both $u$ and $v$ have at least one non-zero component. Then $uv^T$ is an $n \times n$ matrix with at least one non-zero component. Moreover, let us find the 2x2 sub-matrix of $uv^T$ in rows $r,\rho$ and columns $c,\gamma$. Letting $u_r=a, u_{\rho}=b,v_c=e,v_{\gamma}=f$ it is $$\begin{bmatrix} ae&af\\be&bf\\\end{bmatrix}$$ whose determinant is 0. Thus the determinant of every 2x2 sub-matrix of $uv^T$ is 0, so the rank of $uv^T$ is 1 and the dimension of its solution-space is $n-1 \ge 1.$ Look at the rows of $uv^T$.They are just the rows of $v^T$ multiplied by $u_1,...,u_n$. Bu at least one of $u_1,...,u_n$ is non-zero, say $u_{u_{i*}}$ so writing the condition $uv^Tx=0$ as a set of $n$ linear equations we can take out the factor $u_{i*}$ from the $i*$ -th equation, giving us the equaion $v^Tx=0.$