Let $V$ be a real vector space of finite dimension $n$. Let $v,v'\in V$ be linearly independent vectors and $w,w'\in V$ also linearly independent vectors. Prove that the tensor $v\otimes w+v'\otimes w'$ cannot be equal to a pure tensor, i.e. a tensor of the form $a\otimes b$, with $a, b\in V$.
MY IDEA: suppose that the thesis is false, i.e. suppose that there exist vectors $a, b\in V$ such that $v\otimes w+v'\otimes w'=a\otimes b$. This means that $\forall \varphi,\psi\in V^* $ we have $\varphi(v)\psi(w)+\varphi(v')\psi(w')=\varphi(a)\psi(b)$. Now I don't how to proceed.
I'm going to change notation for a reason that will soon be obvious. Let $v=v_1$, $v'=v_2$. Since $v_1,v_2$ are linearly independent, we can extend to a basis $\{v_1,\dots,v_n\}$. Let $\{\phi^1,\dots,\phi^n\}$ be the dual basis. Do the same with $\{w_i\}$ and $\{\psi^i\}$. Write $a=\sum c_iv_i$ and $b=\sum d_iw_i$.
Then, taking $\phi=\phi^i$, $\psi=\psi^j$, we obtain the following: \begin{alignat*}{2} i=j=1 &\qquad&c_1d_1=&1 \\ i=j=2 && c_2d_2=&1 \\ i=1, j=2 && c_1d_2=&0 \\ \end{alignat*} This is an immediate contradiction, so no such $a,b$ can exist.