# $f_1 \circ f_2 \circ f_3=0$

• $$E$$ is a vector space
• $$dim E=3$$
• $$\forall 1 \leq i,j \leq 3$$ :
• $$f_i ^2=0$$
• $$f_i \circ f_j=f_j \circ f_i$$

We want to prove that $$f_1 \circ f_2 \circ f_3=0$$

My attempt :

$$f_1^2=0$$ so $$f_1(Im f_1)=0$$ so $$Im f_1 \subset Ker f_1$$, but $$rg f_1 + dim Ker f_1=3$$ so $$rg f_1 \leq 1$$ and $$rg(f_1 \circ f_2 \circ f_3) \leq rg f_1 \leq 1$$

• If $f_1 f_2$ were zero, you’d be done. If it’s nonzero, the image is in the image of $f_1$and $f_2$ so they’re the same vector, but $f_1^2$ is 0, so $f_1 f_2$ is 0
– Eric
Oct 5 '21 at 21:50

By what you already did we can now say: Either one $$f_i=0$$, then we are done. Or all three $$f_1,f_2,f_3$$ have rank $$1$$. This means that the images of $$f_1,f_2,f_3$$ are given by vectors $$v_1,v_2,v_3$$. Then $$f_i(x) = (a_1^ix_1+a_2^ix_2+a_3^ix_3)v_i = (a^i\cdot x)v_i$$ for some $$a^i$$. Then we get: $$f_i(f_j(x)) = (a^i\cdot((a^j\cdot x)v_j))v_i=(a^j\cdot x) (a^i\cdot v_j)v_i = f_j(f_i(x)) = (a^i\cdot x)(a^j\cdot v_i)v_j$$ In particular this means that all $$v_i$$ are dependent. Also this implies that all $$a^i$$ are dependent (since this holds for all $$x$$). But this means that up to scaling all $$f_i$$ are the same, thus $$f_i\circ f_j = Cf_i^2 =0$$.