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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?

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  • $\begingroup$ Something like (0 1 \\ 0 0) Satisfies T^2 = 0 but T is not zero. Does that help? $\endgroup$ – CJD Jun 2 '18 at 3:51
  • $\begingroup$ Any such operator such that $T\neq 0$ but $T^2=0$ will be such that $\emptyset \subsetneq \text{Image}(T)\subseteq \ker(T)$. Assuming $V$ is finite dimensional as well, the rank-nullity theorem will imply even more. $\endgroup$ – JMoravitz Jun 2 '18 at 3:55
  • $\begingroup$ @JMoravitz that's what I wanted to know.. thx.. but how to prove that property? $\endgroup$ – Beverlie Jun 2 '18 at 3:56
  • $\begingroup$ Remember what it means to be in the image of an operator. Next, notice that $Tv\in \text{Image}(T)$ for every $v$ and notice that $T(Tv)=T^2v=0v=0$. Now., formalize each of these observations and word the appropriate conclusion. $\endgroup$ – JMoravitz Jun 2 '18 at 3:58
  • $\begingroup$ @JMoravitz you just proved it. thx. Any hint for me to proceed further to deal with $H$? What kind of the set $H$ would imply to show that entire vector space would be seperated into different those 3 sets? $\endgroup$ – Beverlie Jun 2 '18 at 4:01
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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

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

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  • $\begingroup$ $\operatorname{Im} T^t \,\dot+\, \ker T = V$ Why is it obvious? $\endgroup$ – Beverlie Jun 2 '18 at 9:32
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    $\begingroup$ @Beverlie It isn't. I proved it in the next two lines. $\endgroup$ – mechanodroid Jun 2 '18 at 9:33
  • $\begingroup$ V=ImTt+˙kerT⊆kerTt+˙kerT Is oppsite side of inclusion trivial? $\endgroup$ – Beverlie Jun 2 '18 at 9:36
  • $\begingroup$ @Beverlie Yes. $\ker T \subseteq V$ and $\ker T^t \subseteq V$ implies $\ker T +\ker T^t \subseteq V$. $\endgroup$ – mechanodroid Jun 2 '18 at 9:40
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    $\begingroup$ @Beverlie Yes. The first direct sum can also be proved like this: assume $T^tx + y = 0$ for some $x \in V, y \in \ker T$. Apply $T$ to obtain $$0 = TT^tx + Ty = TT^tx$$ and now conclude $T^tx = 0$ like above. Then also $y = 0$. Therefore, the sum $\operatorname{Im} T^t + \ker T$ is direct. In both cases you still need to check dimensions to prove that the sum is equal to $V$. $\endgroup$ – mechanodroid Jun 2 '18 at 11:33

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