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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

  1. $\mathrm{im} S = \mathrm{im} T = V$,
  2. $\ker S^\infty \cap \ker T^\infty = \{ 0 \}$,
  3. $\dim \ker S < \infty$ and $\dim \ker T < \infty$,
  4. $ST = TS$

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.

Eric Wofsey
  • 330,363
Max K
  • 93

1 Answers1

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It suffices to assume that $ST=TS$, $\ker T$ is finite dimensional, and $\ker S\cap \ker T=0$. Given these assumptions, suppose $v\in \ker(ST)$. 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 $n>0$, $B\subseteq \ker T$. 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 $v\in B$ and hence $v\in\ker T$. Thus we may assume the constant term of $p$ is $0$ and write $p(x)=xq(x)$ for some polynomial $q$.

Observe now that $q(S)v\in\ker S$, and is nonzero by the minimality of $p$. If the constant term of $q$ is zero, then $q(S)v\in B\subseteq \ker T$, but that is impossible since $\ker S\cap \ker T=0$. Thus $q$ has nonzero constant term, and we may multiply by a scalar to assume the constant term is $1$. But then $v-q(S)v\in B\subseteq \ker T$, and so $v=q(S)v+(v-q(S)v)\in\ker S+\ker T$.

Eric Wofsey
  • 330,363
  • Thank you, excellent answer, and I don't know if I would have thought of that myself. – Max K Jul 24 '20 at 00:05
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    For what it's worth, I came at it from a perspective of commutative algebra rather than linear algebra. A vector space with two commuting operators is just a $k[x,y]$-module, and focusing on the submodule generated by an element $v\in\ker(ST)$ means looking at a cyclic $k[x,y]/(xy)$-module, and then assuming $\ker S$ and $\ker T$ are finite-dimensional means that this module is finite-dimensional over $k$. – Eric Wofsey Jul 24 '20 at 00:18