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  • $E$ 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 $rg(f_1 \circ f_2 \circ f_3) \leq rg f_1 \leq 1$

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  • $\begingroup$ 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 $\endgroup$
    – Eric
    Oct 5 '21 at 21:50
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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 $f_i\circ f_j = Cf_i^2 =0$.

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