$\mathbb R$-forms of $V^2$ i.e. subspaces that are isomorphic to $V$

Question

Let $V$ be an $\mathbb R$-vector space. Let $A$ be an $\mathbb R$-subspace of $V^2$ that is $\mathbb R$-isomorphic to $V$. Then $A$ could be $V \times 0$ or $0 \times V$. Is every other $A$ given by $\{(v,b_Av)|v \in V\}$ for some unique $b_A \in \mathbb R \setminus 0$?

Context:

$f$ is the complexification of a map if $f$ commutes with almost complex structure and standard conjugation. What if we had anti-commutation instead?

Complexification of a map under nonstandard complexifications of vector spaces

Answer 1

No, of course not. See for instance $V=\Bbb R^2$ and $$A=\{(x_1,x_2,x_3,x_4)\in V\times V\,:\, x_2=0\land x_3=0\}.$$