What I did:
Let $f_1,f_2,f_3: \mathbb{R}³ \rightarrow \mathbb{R}³$ linear operators and take $\exists y\in\mathbb{R}³,y\neq0 $. Let $\alpha, \beta, \gamma \in \mathbb{R}$ such that $\alpha f_1(y) + \beta f_2(y) + \gamma f_3 (y) = 0$. But, we can associate the linear operator $f_i$ with a bilinear form $B_i: \mathbb{R}³ \times \mathbb{R}³ \rightarrow \mathbb{R}$ such as $B_i(x,y) = <f_i(y),x>$. So, $\forall x \in \mathbb{R}³$ $$ 0 = < \alpha f_1(y) + \beta f_2(y) + \gamma f_3 (y), x > = \alpha <f_1(y),x> + \beta <f_2(y),x> + \gamma <f_3 (y),x> $$ $$0 = \alpha B_1 (x,y) + \beta B_ 2(x,y) + \gamma B_3 (x,y)$$
What I also know:
Well, if I proof that $B_1(x,y),B_2(x,y),B_3(x,y)$ are l.d. I solve the problem. Besides, $\bigcap_{i=1}^³ \ker B_i =${$(x,0),(0,x): x \in \mathbb{R}^3$} if, and only if, $f_1(y),f_2(y),f_3(y)$ are l.i., $\forall y \in \mathbb{R}^3,y \neq 0$.
A linear operator $f$ on $R^3$ has a real-valued eigenvalue $x$, because if you represent $f$ by a matrix with respect to any basis for $R^3$, the equation $\det (f-xI)=0$ is a real cubic. Now if $f_1$ is invertible, let $x$ be an eigenvalue of the operator $g=f_1^{-1}(f_2+f_3)$ and let $g(y)=xy$ with $y \ne 0$. Then $(-x)f_1(y)+f_2(y)+f_3(y)=0$. But if $f_1$ is not invertible let $f_1(y)=0 \ne y$ so the set $\{ {f_1(y),f_2(y),f_3(y)} \}$ contains $0$ so it's not l.i.