# On a special kind of $6$-dimensional vector subspace of $\mathbb C^9$

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

$$V=\{(x_1,...,x_9)\in \mathbb C^9 : (x_3,x_6,x_9)^T = A (x_1,x_2,x_4,x_5,x_7,x_8)^T \}=\{(x_1,...,x_9)\in \mathbb C^9 : (x_3,x_6,x_9)^T = B(x_1,x_2,x_4,x_5,x_7,x_8)^T \}$$

; then is it true that $$A=B$$ ?

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 indent. 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 $$A=B$$.