If $\sigma:V\to \mathbb{R}$, with $V$ a vector space and $\sigma(v)=\phi_1(v)\phi_2(v) $ for $\phi_1,\phi_2\in V^{*}$ then $\phi_1=0$ or $\phi_2=0.$ If $\sigma:V\to \mathbb{R}$, where $V$ is a vector space  and $\sigma(v)=\phi_1(v)\phi_2(v) $ for some $\phi_1,\phi_2\in V^{*}.$ Show that $\phi_1=0$ or $\phi_2=0.$ Note that $\sigma \in V^{*}.$
My Attempt: Suppose $\phi_1$ and $\phi_2$ are not equal to $0.$ Then there exists $x,y\in V$ such that $\phi_1(x)=a\not =0$ and $\phi_2(y)=b\not =0.$ So we have that $\sigma(x)=a\phi_2(x)=ka^2$ for some $k\in \mathbb{R}.$ And similarily $\sigma(y)=\gamma b^2$ for some $\gamma\in \mathbb{R}.$ Thus $$\sigma(x+y)=(\phi_1(x)+\phi_1(y))(\phi_2(x)+\phi_2(y))=(a+\gamma b)(ka+b)$$ $$\implies ka^2+ab+k\gamma ab+\gamma b^2=ka^2+\gamma b^2.$$ This means that $ab=0$ or $k\gamma =-1.$ Since $ab\not =0$ if must be that $k\gamma=-1.$ I am not sure how to proceed after this. 
 A: First prove that $\sigma=0$. Let $v\in V$ and $\sigma(v)=z$, then $\sigma(2v)=2\sigma(v)=2z$ and also $\sigma(2v)=\varphi_1(2v)\varphi_2(2v)=4\varphi_1(v)\varphi_2(v)=4\sigma(v)=4z$, thus you have $2z=4z$ i.e. $z=0$.
Now, suppose $\varphi_1\ne 0$ and $\varphi_2\ne 0$. Thus, there are $x,y\in V$ such that $\varphi_1(x)=a\ne 0$ and $\varphi_2(y)=b\ne 0$. However, due to $\sigma(x)=\sigma(y)=0$ we have $\varphi_2(x)=0$ and $\varphi_1(y)=0$. Now, let us calculate $\sigma(x+y)$: $0=\sigma(x+y)=(\varphi_1(x)+\varphi_1(y))(\varphi_2(x)+\varphi_2(y))=\varphi_1(x)\varphi_2(y)=ab$. In other words, $ab=0$, which is a contradiction.
A: Once it is shown that $\sigma=0$ (as done by user8734617) i.e., $\ker\sigma=V$, then note that $\ker\sigma=\ker\phi_1\cup\ker\phi_2$. That is, the union of two subspaces of $V$ is again a subspace of $V$ and therefore either $\ker\phi_1\subseteq \ker\phi_2$ or $\ker\phi_2\subseteq\ker\phi_1$. So $\ker\sigma=\ker\phi_1$ or $\ker\sigma=\ker\phi_2$ i.e., either $\ker\phi_1=V$ or $\ker\phi_2=V$ and so $\phi_1=0$ or $\phi_2=0$.
