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Prove: $T \in L(V,V), T^2 = 0 \iff T(v) \subset n(T)$

Is the following correct?

Proof: $\rightarrow$

Let $T^2 = 0 \iff T(T(v)) = 0$

Suppose $x \in T(v)$ we must show that $x \in n(T) \iff T(x) = 0$

$x\in T(v)$ implies there exists $v\in V$ s.t. $T(v)=x$

Consider $T(T(v))$:

$T(T(v)) = T(x) = 0$

$\leftarrow$ Let $T(v) \subset n(T)$ Suppose $x\in T(v)$. Then we know $x\in n(T) \iff T(x) = 0$. We also know there exists $v\in V$ so that $T(v) = x$.

Consider $T(T(v))$:

$T(T(v)) = T(x) = 0$

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How should "Let $T^2 = 0 \iff T(T(v)) = 0$" be interpreted? $\:$ Should it be "If $T(T(v)) = 0$ then $\hspace{.8 in}$ redefine $T$ to be such that $T^2 = 0$, else leave $T$ as it was." ? $\;\;$ –  Ricky Demer Nov 14 '12 at 3:00
Also, the elements of $T(v)$ should be completely irrelevant. $\:$ (On the other hand, $\hspace{1.8 in}$ the elements of $\operatorname{Range}(T)$ should be important.) $\;\;$ –  Ricky Demer Nov 14 '12 at 3:05

1 Answer 1

up vote 2 down vote accepted

Your proof is correct in spirit but it lacks clarity. You need to take care in distinguishing $T(V)$ which is a set (the image of $T$) and $T(\mathbf{v})$ which is a single vector (the image of $\mathbf{v}$ under $T$).

I will rewrite your proof, following your exact logic. Hopefully you will find this version more clear.

Theorem: Let $T:\ V\rightarrow V$ be a linear mapping. Then $T^2 = 0 \iff T(V) \subseteq \operatorname{null}(T)$.

Proof: ($\Rightarrow$) Suppose that $T^2 = 0$. Then for all $\mathbf{v} \in V$ we have $$T^2(\mathbf{v}) = T(T(\mathbf{v})) = \mathbf{0}$$ This shows that $T(\mathbf{v}) \in \operatorname{null}(T)$ for all $\mathbf{v} \in V$ and therefore $T(V) \subseteq \operatorname{null}(T)$.

($\Leftarrow$) Conversely, suppose that $T(V) \subseteq \operatorname{null}(T)$. This means that for all $\mathbf{v} \in V$ we have $$T(\mathbf{v}) \in \operatorname{null}(T) \implies T(T(\mathbf{v})) = \mathbf{0}$$ Therefore $T^2 = 0$. $\square$

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