Exercise 13, Section 3.1 of Hoffman’s Linear Algebra

Question

Let $V$ be a vector space and $T$ a linear transformation from $V$ into $V$. Prove that following two statements about $T$ are equivalent.

$(a)$ The intersection of the range of $T$ and the null space of $T$ is the zero subspace of $V$.

$(b)$ If $T(T(\alpha))=0$, then $T(\alpha)=0$.

My attempt: $(a)\Rightarrow (b)$. Suppose $R_T\cap N_T=\{0_V\}$. If $T(T(\alpha))=0_V$, for some $\alpha \in V$, then $T(\alpha)\in N_T$. Since $T:V\to V$, we have $T(\alpha)\in R_T=T(V)$. Thus $T(\alpha)\in N_T \cap R_T=\{0_V\}$. Hence $T(\alpha)=0_V$.

$(a)\Leftarrow (b)$. Let $u\in R_T\cap N_T$. Then $u\in R_T$ and $u\in N_T$. So $\exists x\in V$ such that $T(x)=u$, and $T(u)=0_V$. Since $T(x)\in V$, we have $T(T(x))=T(u)=0_V$. By hypothesis, $T(x)=0_V=u$. Thus $u\in \{0_V\}$. Hence $N_T \cap R_T=\{0_V\}$.

Is my proof correct? In Proof we didn’t use $T$ is a linear map. Wait! Perhaps we did, because without linearity condition, $N_T$ could be empty and/or $N_T\cap \{0_V\}=R_T\cap \{0_V\}=\emptyset$. Am I right?

Edit: Comment by Arturo Magidin:

Linearity of $T$ is not strictly necessary: you just need the range and kernels to have a nonempty intersection (the result holds for group homomorphisms, for example). If you do not assume that they have a nonempty intersection, your second part proves that the intersection is contained in $\{0_V\}$ , and hence is either that single vector, or is empty.

$T$ is a linear map$\implies$$R_T \cap N_T \neq \emptyset$, since $0_V\in R_T \cap N_T$.

Answer 1

Your proof is correct and reasoning also.

Where we use linearity of $T$ ?

•If $T$ is linear map then $T 0=0$ implies both $N_T $ and $R_T$ contains zero vector (both are non empty at least!) . That allows you to choose $u\in N_T \cap R_T$ .