# Prove that the image of the bilinear application is not a vector space

Prove that the image of the bilinear application $$\mathbb{R}^2\times\mathbb{R}^2 \ni \big((x_1,x_2),(y_1,y_2)\big) \mapsto \varphi \big((x_1,x_2),(y_1,y_2)\big) = (x_1y_1,x_1y_2,x_2y_1,x_2y_2)\in\mathbb{R}^4$$ is not a vector space.

My work. If $$\mathrm{Im}(\varphi)\subseteq\mathbb{R}^4$$ is a subspace of $$\mathbb{R}^{4}$$ then for any scalars $$\mu,\lambda\in\mathbb{R}$$ and vectors $$p=(a\cdot c, a\cdot d, b\cdot c, b\cdot d) \qquad \mbox{ and } \qquad q=(u\cdot z, u \cdot w, v\cdot z, v\cdot w)$$ in $$\mathrm{Im}(\varphi)$$ we have $$\mu\cdot p+\lambda\cdot q \in \mathrm{Im}(\varphi).$$ Therefore, if we can show that there are scalars $$\lambda,\mu\in\mathbb{R}$$ and vectors $$p,q\in \mathrm{Im}(\varphi)$$ such that $$\mu\cdot p+\lambda\cdot q \notin \mathrm{Im}(\varphi)$$ then $$\mathrm{Im}(\varphi)$$ will not be a vector space. The relationship of non-pertinence above is equivalent to the following statement. For all vector $$(x_1y_2,x_1y_2,x_2y_2,x_2y_2)\in\mathrm{Im}\varphi$$ the system below has no solution $$\begin{array}{rcl} \mu a c+ \lambda u z &=& x_1y_1\\ \mu a d+ \lambda u w &=& x_1y_2\\ \mu b c+ \lambda v z &=& x_2y_1\\ \mu b d+ \lambda v w &=& x_2y_2\\ \end{array}$$ But I have been unable to find real numbers $$a$$, $$b$$, $$c$$ and $$d$$ and scalars $$\lambda$$ and $$\mu$$ for which the above system has no solution for $$x_1$$, $$x_2$$, $$y_1$$ and $$y_2$$ varying in $$\mathbb{R}$$.

Your claim is true as $$\,\operatorname{Im}\varphi\,$$ is $$\,$$ not closed $$\,$$ under addition:
$$\varphi\big((1,0),(1,0)\big) \:=\: (1,0,0,0)\\ \varphi\big((0,1),(0,1)\big) \:=\: (0,0,0,1)$$
Now consider the sum $$(1,0,0,1)$$ of both ...
Can it be contained in $$\operatorname{Im}\varphi$$?