let V be a vector space, T : V → V be a linear transformation. Prove that $T^2 = 0$ if and only if $R(T) \subset N(T)$.

Question

Attempt:

$T^2 = 0$ if and only if for all $v \in V$ , $0 = T^2(v) = T(T(v))$ if and only if for all $v \in V , T(v) \in N(T)$ if and only if $R(T) \in N(T)$.

If the question were this instead: Let $T : V \to V$ be a linear operator for a vector space $V$. Prove that $R(T) \subset N(T)$ if and only if $T^2 = 0$, would the proof above ^ stil be fine ?

Answer 1

I can't see any logical distinction 'twixt the two, so both proofs look fine to me.

That being said, I would likely frame the argument something like this:

Given that

$T^2 = 0, \tag 1$

suppose

$x \in R(T); \tag 2$

then

$\exists y \in V \mid x = T(y); \tag 3$

thus

$T(x) = T^2(y) = 0 \Longrightarrow x \in N(T); \tag 4$

thus we see that

$R(T) \subset N(T); \tag 5$

now assuming (5), we have

$\forall y \in V \; T^2(y) = T(T(y)); \tag 6$

since

$T(y) \in R(T), \tag 7$

from (5) we obtain

$\forall y \in V, \; T^2(y) = T(T(y)) = 0, \tag 7$

whence

$T^2 = 0; \tag 8$