Let $V \subseteq \mathbb C^9$ be a vector subspace of dimension $6$. Suppose that there exists $A,B \in M_{3 \times 6} (\mathbb C)$ such that
$$V=\{(x_1,\ldots,x_9)\in \mathbb C^9 : (x_3,x_6,x_9)^T = A (x_1,x_2,x_4,x_5,x_7,x_8)^T \}=\{(x_1,\ldots,x_9)\in \mathbb C^9 : (x_3,x_6,x_9)^T = B(x_1,x_2,x_4,x_5,x_7,x_8)^T \}.$$
Then, is it true that $A=B$?
Best Answer
Yes, this is true. Up to the numbering of coordinates, you just write $V=\{(y,Ay):y\in\mathbb C^6\}$ and $V=\{(y,By):y\in\mathbb C^6\}$. Now if you insert the basis vectors $e_1,\dots,e_6$, you readily see that the vectors $(e_1,Ae_1),\dots,(e_6,Ae_6)$ are linearly independent. Since $\dim(V)=6$, they have to be a basis of $V$. But this implies that any element $v\in V$ is determined by the coordinates $v_1,\dots,v_6$, so given $y\in\mathbb C^6$, there is at most one $z\in\mathbb C^3$ such that $(y,z)\in V$, and this implies $A=B$.