8
$\begingroup$

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!

$\endgroup$
2
  • $\begingroup$ Should even be $\oplus.$ $\endgroup$ Jun 28 '18 at 16:02
  • $\begingroup$ @Chickenmancer, not if $f=g$. $\endgroup$
    – lhf
    Jun 28 '18 at 17:42
1
$\begingroup$

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 $1$.

$\endgroup$
2
  • 1
    $\begingroup$ Do we have $fg=gf$ here? $\endgroup$ Jul 7 '18 at 19:46
  • $\begingroup$ @PeterFranek: You are right. I did not consider this assumption. $\endgroup$ Jul 7 '18 at 19:57
1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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 '18 at 8:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.