Take two linear transformations $T \colon U \to V$ and $S \colon V \to W$ where $U$, $V$ and $W$ are finite. I want to show that $$ \dim \ker (S \circ T) \leq \dim \ker S + \dim \ker T. $$

Attempt: I've been using the dimensional theorem because of the finity from each vector space. So I got that \begin{align*} &\, \dim \ker T + \dim \ker S \\ =&\, \dim \ker (S \circ T) + \dim V + \dim \operatorname{im} (S \circ T) - \dim \operatorname{im} T - \dim \operatorname{im} S. \end{align*} I have been using the inequalities

  1. $\dim \operatorname{im} S > \dim \operatorname{im} (S \circ T)$,
  2. $\dim V > \dim \mathrm{im} T$,

and that $\ker T \subseteq \ker (S \circ T)$.

But with such inequalities I couldn’t conclude anything. Any tips or hint in order to progress or get the answer?


1 Answer 1


We do it like this:

Lemma $1$: $T[\ker (S \circ T)] = \ker S \cap {\rm Im}\,T$.

Proof: you do it, okay?

Lemma $2$: If $F\colon V_1 \to V_2$ is linear and $Z \subseteq V_2$ is a subspace, then $F^{-1}[Z]$ is a subspace of $V_1$ and $\dim F^{-1}[Z] \leq \dim \ker F + \dim Z$.

Proof: I'll check only the formula. Applying the rank-nullity theorem for the restriction of $F$ to $F^{-1}[Z]$, we get $$\dim F^{-1}[Z]= \dim \ker F\big|_{F^{-1}[Z]}+\dim {\rm Im}\,F\big|_{F^{-1}[Z]} \leq \dim \ker F + \dim Z.$$

Now to prove what you want we need only note that

$$\begin{align} \dim \ker(S \circ T) &\stackrel{(1)}{\leq} \dim T^{-1}[T[\ker S \circ T]] \\ &\stackrel{(2)}{=} \dim T[\ker(S \circ T)] + \dim \ker T \\ &\stackrel{(3)}{=} \dim (\ker S \cap {\rm Im}\,T)+\dim \ker T \\ &\stackrel{(4)}{\leq} \dim \ker S + \dim \ker T,\end{align}$$where in (2) we apply lemma $2$, in (3) we apply lemma $1$, and (1) and (4) follow because of elementary set-theoretic considerations.

  • $\begingroup$ Very nice answer (+1). $\endgroup$ Oct 25, 2018 at 13:23

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