linearly dependent linear tranformations Let $E$ be a vector space over $k$ of finite dimension $n$. Let $f_1, \ldots, f_p \in \mathcal{L}(E)$ with $p \leq n$ and assume that the $f_i$ are simultaneously diagonalizable (put it simply, there exists a basis of $E$, where the matrices representing the $f_i$ are all diagonal).
Assume that for all $x \in E$, the vectors $f_1(x), \ldots, f_p(x)$ are linearly dependent. Then the vectors $f_1, \ldots, f_p$ are linearly dependent in $\mathcal{L}(E)$.
I have a complicated proof of that, and I'd like to know if there is a simple argument proving this result. Furthermore, I feel like the diagonalizability assumption could be withdrawn and one could only assume that for any $i,j$ we have $f_i \circ f_j = f_j \circ f_i$, but I have no clue on how to prove that last assertion.
 A: Without loss of generality $E=k^n$ and the $f_i$ are identified with diagonal matrices $D_i$ with $f_i\colon x\mapsto D_i x$. Now note that for $D=\operatorname{diag}(\lambda_1,\dots,\lambda_n)$ and $\mathbf 1=(1,\dots,1)^t$ we have $D\mathbf 1 = (\lambda_1,\dots,\lambda_n)^t$.
So taking $x=\mathbf 1$ you get that the vectors $D_i\mathbf 1$, which consist of the diagonal entries of the $D_i$, are linearly dependent. Then of course the corresponding diagonal matrices satisfy the same linear relation.
A: Counterexample for the case when the matrices are not simultaneously diagonalisable:
$$f_1:=\begin{pmatrix}1&1\\0&0\end{pmatrix},\qquad f_2:=\begin{pmatrix}1&2\\0&0\end{pmatrix}$$
Then $f_1,f_2$ are linearly independent, yet for any vector $x=(a,b)^t$, $$\tfrac{1}{a+b}f_1(x)+\tfrac{1}{a+2b}f_2(x)=0$$ (unless $a+b=0$ or $a+2b=0$; those cases correspond to eigenvectors of $f_1$, $f_2$ respectively, so $f_1(x)=0$ or $f_2(x)=0$, so $f_1(x)$, $f_2(x)$ are still dependent.)

Looking at the $2\times2$ case, commutativity of the matrices is not required for the property to hold; rather it is invertibility.
Suppose $f_1=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ is invertible and $f_2=\begin{pmatrix}\alpha a&\beta b\\\alpha c&\beta d\end{pmatrix}$ are as in the question. The second matrix must be of this form since $f_1(\mathbf{i})$, $f_2(\mathbf{i})$ are linearly dependent, and similarly $f_1(\mathbf{j})$, $f_2(\mathbf{j})$. Furthermore, $f_1(x)$ and $f_2(x)$ are in the same direction, hence if $x=(u,v)^t$, then $$\lambda\begin{pmatrix}au+bv\\cu+dv\end{pmatrix}=\begin{pmatrix}\alpha au+\beta bv\\\alpha cu+\beta dv\end{pmatrix}$$ $$\therefore\ (au+bv)(\alpha cu+\beta dv)=(cu+dv)(\alpha au+\beta bv)$$ $$\therefore\ (\beta-\alpha)(ad-bc)uv=0$$ Hence $\alpha=\beta$, that is, $f_2=\alpha f_1$ as required.
