# Question about Image and Null space of a linear operator and invariant subspace.

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

This question has asked before here:

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 $$\ker T \cap Im T = \{0\}??$$

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:
$$\dim(\ker T \cap \operatorname{Im} T) = \dim\ker T + \dim \operatorname{Im} T - \dim(\ker T + \operatorname{Im} T) = \dim V-\dim V = 0.$$
• If $W$ and $\text{im }T$ are complementary subspaces, do we necessary have that their intersection is the zero vector? Jan 26, 2021 at 4:41
• 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 Jan 26, 2021 at 14:06