Let $T:V \to V$ a linear operator of finite dimensional vector space $V$. Show that $ImT$ has a complementary $T-$invariant subspace $\iff$ $Im T \cap \ker T = \{0\}$.

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Cyclic Decompositions and the Rational form

I got the side $(\Leftarrow)$, but I can't understand the otherside. Let me explain my question. If $W$ is a $T-$invariant subspace of $ImT$, of course that given $w \in W$ then $T(w) \in \ker T$, then $W \subset \ker T$. Ok, why it implies that $\ker T \cap Im T = \{0\}??$


1 Answer 1


Assume that $V= W \dotplus \operatorname{Im} T$ where $W$ is $T$-invariant. We claim that $W \subseteq \ker T$.

If $w \in W$ then $Tw \in W$ as well, but also $Tw \in \operatorname{Im} T$ and hence $w \in W\cap \operatorname{Im} T=\{0\}$. Therefore $Tw = 0$ so $w \in \ker T$.

Now we have $$V= W \dotplus \operatorname{Im} T \le \ker T + \operatorname{Im} T$$ so $\ker T + \operatorname{Im} T = V$. It remains to show that the sum is direct, but this follows directly from the rank-nullity theorem:

$$\dim(\ker T \cap \operatorname{Im} T) = \dim\ker T + \dim \operatorname{Im} T - \dim(\ker T + \operatorname{Im} T) = \dim V-\dim V = 0.$$

  • $\begingroup$ If $W$ and $\text{im }T$ are complementary subspaces, do we necessary have that their intersection is the zero vector? $\endgroup$
    – Matthew H.
    Jan 26, 2021 at 4:41
  • $\begingroup$ @MatthewPilling Always. See this. $\endgroup$
    – azif00
    Jan 26, 2021 at 5:30
  • $\begingroup$ When I was working on this problem I googled "complementary subspace" and it claimed $U$ and $V$ are complementary iff $U+V$ is the entire vector space. It didn't specify that the sum must be direct. Good to know $\endgroup$
    – Matthew H.
    Jan 26, 2021 at 14:06

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