# 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 Furthermore, I feel like the diagonalizability assumption could be withdrawn and one could only assume that for any $$i,j$$ we have $$f_i \circ f_j = f_j \circ f_i$$, but I have no clue on how to prove that last assertion.

• I don't have a reference on hand, but I'm pretty sure the $f_{i}$ will all pairwise commute if and only if they are simultaneously diagonalizable Nov 23, 2020 at 10:05
• @MorganRodgers : Of course, if you assume that all the $f_i$ pairwise commute and are diagonalizable, then they are simultaneously diagonalizable. In my second question, I withdraw the assumption that they are diagonalizable and only assume that they pairwise commute. Nov 23, 2020 at 17:52

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.

• Thanks! Indeed that was so simple, I feel ashame I didn't find this neat argument. Nov 23, 2020 at 17:49

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.

• Thanks, nice example! Though here $f_1$ and $f_2$ do not commute. I might be mistaken, but I have the feeling that the hypothesis $f_1$ and $f_2$ commutes is the key hypothesis (not the invertibility one). Nov 23, 2020 at 17:49
• @Libli When you analyse the $2\times2$ case, the determinant comes out as a factor. So two invertible $2\times2$ matrices have the required property without commuting. Nov 24, 2020 at 5:25
• that is very interesting. Could you ellaborate a bit on what you mean by "the determinant comes out as a factor"? Nov 24, 2020 at 8:06
• @Libli Added the requested part to the answer. Nov 24, 2020 at 10:25
• This is very interesting thanks. It seems your proof can be restated as follows. Let $f$ and $g$ such that $f(x)$ and $g(x)$ ae linearly dependant for all $x \in E$. Since $f$ is invertible, one sees that $x$ and $f^{-1}(g(x))$ are linearly dependent for all $x \in E$. Then by Schur's lemma, $\mathrm{id}$ and $f^{-1} \circ g$ are linarly dependent in $\mathcal{L}(E)$, which proves that $f$ and $g$ are linearly dependent in $\mathcal{L}(E)$. Nov 24, 2020 at 13:01