If $T : V \to V$ is linear transformation that satisfies $T\circ T=0$, which statement is true?

Question

If $T: V \to V$ is a linear transformation of vector space $V$ and $T\circ T=0$, then:

a) $\ker T \subseteq \operatorname{image} T$

b) $\operatorname{image} T \subseteq \ker T$

c) $T = 0$

d) $T$ is a nonsingular linear transformation.

My Try: As $T(T(x))=0$ that means $\operatorname{image} T \subseteq \ker T$. But kernel and image are always disjoint, except $0$ is always common in them. That means the only possibility is $T(x)=0$ for every $x\in V$. So it is a zero linear transformation.

But on the answer sheet, option (d) is given as correct.

Answer 1

The image and kernel are not always transverse; consider for example $$T : V \to V, \qquad T(v, w) := (w, 0) .$$

Certainly (d) cannot be correct, as it implies $\det (T \circ T) = (\det T)^2 \neq 0$.