$V = \operatorname{Im} T \oplus \operatorname{Im}T^t \oplus H$ where $T^2 = 0$ and $H = \ker T \cap \ker T^t$

Question

Let $T$ be a square matrix regarded as a linear operator on a finite dimensional vector space $V$ such that $T^2 = 0$. Let $H = \ker T \cap \ker T^t$. Show that $V = \operatorname{Im} T \oplus \operatorname{Im}T^t \oplus H$.

---

[my proof]

If $T = 0$, $V = \operatorname{Im} T \oplus \operatorname{Im}T^t \oplus H = \{0\} \oplus \{0\} \oplus V = V$

If $T \neq 0$, $V = \ker T\cup \ker T^t$ by rank-nullity theorem (* But I am still confusing how the rank-nullity theorem proves this)

and we know $\operatorname{Im} T\subseteq \ker T$ and $\operatorname{Im} T^t\subseteq \ker T^t$ from $T^2 =0$.

Now we need to prove that $\ker T \setminus \operatorname{Im} T = \emptyset$ and $\ker T^t\setminus \operatorname{Im} T^t = \emptyset$ to finish the proof, but can't know how to prove it.

Any hint?

Answer 1

Note that $\operatorname{Im} T^t \,\dot+\, \ker T = V$. Firstly, the sum is direct (i.e. intersection is trivial):

$$T^tx \in \operatorname{Im}T^t \cap \ker T \implies TT^tx = 0 \implies (T^tx)^t(T^tx) = x^tTT^tx = 0 \implies T^tx = 0$$

and secondly, $\dim\operatorname{Im} T^t + \dim\ker T = \dim\operatorname{Im} T + \dim\ker T = \dim V$, by the rank-nullity theorem.

Now, as you stated, $T^2 = 0$ implies that $\operatorname{Im} T \subseteq \ker T$ and $\operatorname{Im} T^t \subseteq \ker T^t$

$$V = \operatorname{Im} T^t \,\dot+\, \ker T \subseteq \ker T^t \,\dot+\, \ker T$$ so $\ker T^t \,\dot+\, \ker T = V$.

Now we claim that $V = \operatorname{Im} T \,\dot+\, \operatorname{Im} T^t \,\dot+ \,H$.

Firstly, the sum is direct. Assume that $0 = Tx + T^ty + z$ with $x,y \in V, z \in H$. Applying $T$ and $T^t$ gives

$$0 = T^2x + TT^t + Tz = TT^ty \implies (T^ty)^tT^ty = y^tTT^ty = 0 \implies T^ty = 0$$ $$0 = T^tTx + (T^t)^2z + Tz = TT^ty \implies (Tx)^t(Tx) = x^tT^tTx = 0 \implies Tx =0$$

and thus also $z = 0$.

Secondly, we have:

\begin{align} \dim H &= \dim (\ker T \cap \ker T^t) \\ &= \dim\ker T + \dim\ker T^t - \dim(\ker T\,\dot+\, \ker T^t) \\ &= \dim\ker T + \dim\ker T^t - \dim V \end{align}

so \begin{align} \dim\operatorname{Im} T + \dim \operatorname{Im} T^t + \dim H &= \dim\operatorname{Im} T + \dim \operatorname{Im} T^t + \dim\ker T + \dim\ker T^t - \dim V\\ &= 2\dim V - \dim V\\ &= \dim V \end{align}

by the rank-nullity theorem.

Hence we conclude

$$V = \operatorname{Im} T \,\dot+\, \operatorname{Im} T^t \,\dot+ \,H$$