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Let $k$ be a field and $V$ be a finite dimensional $k$-vector space. Let $f$ and $g$ be two $k$-linear endomorphisms of $V$ such that $f\circ g=g\circ f$.

Do we have an isomorphism of $k$-vector spaces $V/(\ker f \cap \ker g) \cong \text{im} f +\text{im} g$ ?

Many thanks!

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  • $\begingroup$ Should even be $\oplus.$ $\endgroup$ Jun 28, 2018 at 16:02
  • $\begingroup$ @Chickenmancer, not if $f=g$. $\endgroup$
    – lhf
    Jun 28, 2018 at 17:42

2 Answers 2

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A counter-example seems to be $$ A=\begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix} $$ and $$ B=\begin{pmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}. $$

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  • $\begingroup$ AB=BA=0, and if (e_1,e_2,e_3) is the canonical basis, imA=ke_1, imB=ke_1, kerA=ke_1+ke_3 and kerB=ke_1+ke_2. It is a counterexample, thanks! $\endgroup$
    – Stabilo
    Jul 13, 2018 at 8:03
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Note: In my answer I did not consider the assumption $f\circ g=g\circ f$.

The claim is wrong. Consider the maps $$f:=\left[\matrix{0&0&0\cr0&0&0\cr 1&1&0\cr}\right],\qquad g:=\left[\matrix{0&0&0\cr0&0&0\cr 1&0&1\cr}\right]\ .$$ Then $${\rm ker}(f)={\rm span}\bigl((1,-1,0),(0,0,1)\bigr),\quad {\rm ker}(g)={\rm span}\bigl((1,0,-1),(0,1,0)\bigr)\ .$$ Therefore ${\rm ker}(f)\cap{\rm ker}(g)={\rm span}\bigl((1,-1,-1)\bigr)$ has dimension $1$, hence $V/\bigl({\rm ker}(f)\cap{\rm ker}(g)\bigr)$ has dimension $2$. On the other hand $${\rm im}(f)={\rm im}(g)={\rm im}(f)+{\rm im}(g)={\rm span}\bigl((0,0,1)\bigr)$$ has dimension $1$.

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    $\begingroup$ Do we have $fg=gf$ here? $\endgroup$ Jul 7, 2018 at 19:46
  • $\begingroup$ @PeterFranek: You are right. I did not consider this assumption. $\endgroup$ Jul 7, 2018 at 19:57

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