$\ker ST=\ker T$

Question

Let $S$ and $T$ be linear maps between vector spaces such that the composition $ST$ makes sense. Clearly, $\ker ST\supseteq \ker T$. The two instances that come to my mind for having an equality in this relation are:

1. $S$ is one-to-one;
2. $S=T^*$ (when dealing with bounded operators of Hilbert spaces.)

While looking at some examples I noticed that $\ker ST=\ker T$ also holds when $S$ and $T$ are Fredholm operators (of an infinite-dimensional Hilbert space) with $ST=K_1+I$ and $TS=K_2+I$ where $K_1, K_2$ are compact operators. I can't think of a proof of it, although I believe it should be fairly elementary (at least when $K_i$ are self-adjoint). Any help is appreciated!

Answer 1

In any Hilbert space (say $H=\ell^2(\mathbb N)$), let $T=I$ and $S=I-P$ where $P$ is a finite-rank projection. Then, as $ST=I-P$ and $\ker ST=\text{ran}\,P$, $$ 0=\ker T\subsetneq PH=\ker ST. $$ Both $S$ and $T$ are selfadjoint and Fredholm.

Answer 2

EDIT: This answer no longer applies because the OP changed the question.

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It is not true. Any linear map between finite dimensional vector spaces is Fredholm, so there are finite dimensional counterexamples.

If you want an infinite dimensional counterexample consider $S,T:\ell^2\to\ell^2$, where $S(a_0,a_1,a_2,\ldots)=(0,a_1,a_2,\ldots)$ and $T$ is the identity map.

Answer 3

If the kernel of $S$ contains a nonzero vector $y$ that is in the range of $T$ (let $x$ denote the pre-image of $y$, so we have ker($S$) $\ni y = Tx$), then $x$ lies in the kernel of $ST$, but not in the kernel of $T$.

If there are no such vectors, then ker($S$) $\cap$ range($T$) = the zero vector space.