Mathematics Asked by Max K on December 18, 2021
If $S$ and $T$ are commuting operators on an infinite dimensional vector space $V$, it is in general true that
$$ker S + ker T subseteq ker(ST),$$
but in general equality does not hold. A simple example is given by $S = T = frac{d}{dx}$ on $C^infty(mathbb{R})$. I am looking for conditions on $S$ and $T$ that will give equality in the above equation, ie:
$$ker S + ker T = ker (ST)$$
Writing $ker T^infty$ for $cup_n ker T^n$, I am currently trying to show that the conditions
imply that $ker S + ker T = ker(ST)$. I think the second condition can be weakened to $ker S^2 cap ker T^2 = { 0 }$, but I have this stronger condition for some operators I am interested in. Any help would be appreciated, thanks.
-edit- I am not confident that all these conditions are necessary.
It suffices to assume that $ST=TS$, $ker T$ is finite dimensional, and $ker Scap ker T=0$. Given these assumptions, suppose $vin ker(ST)$. Let $A$ be the span of ${v,Sv,S^2v,dots}$ and let $B$ be the span of ${Sv,S^2v,dots}$. Since $TS^nv=S^{n-1}STv=0$ for any $n>0$, $Bsubseteq ker T$. Since $ker T$ is finite-dimensional, $B$ must be finite dimensional, and hence so is $A$. Let $p$ be the minimal polynomial of $S$ on $A$. If $p$ has a nonzero constant term, then $vin B$ and hence $vinker T$. Thus we may assume the constant term of $p$ is $0$ and write $p(x)=xq(x)$ for some polynomial $q$.
Observe now that $q(S)vinker S$, and is nonzero by the minimality of $p$. If the constant term of $q$ is zero, then $q(S)vin Bsubseteq ker T$, but that is impossible since $ker Scap ker T=0$. Thus $q$ has nonzero constant term, and we may multiply by a scalar to assume the constant term is $1$. But then $v-q(S)vin Bsubseteq ker T$, and so $v=q(S)v+(v-q(S)v)inker S+ker T$.
Answered by Eric Wofsey on December 18, 2021
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