Is my proof that $\ker(T)\subseteq \{v-T(v) \mid v \in V\}$ correct

Question

Given the following: "Let $V$ be a finite dimensional vector space and let $T:V\to V$ be a linear transformation such that $T^2=T$

To show:

$\ker(T) \subseteq \{v-T(v)\mid v\in V\}$

My proof:

Let $v\in V $.

Suppose $v\in \ker(T)$ $\iff T(v)=\underline{0}$ $\iff T(v)=T(T(v))=\underline{0}$ $\iff T(v)-T(T(v))=\underline{0}$ $\iff T(v-T(v))=\underline{0}$ $\iff v-(T(v))\in \ker(T)$

hence $v\in \ker(T)$$\iff v-(T(v))\in \ker(T)$

and if $v$ being in $\ker(T)$ is equivalent to $v-T(v)$ being in $\ker(T)$ then $v$ must be equal to $v-T(V)$ (is this deduction here correct? The logic used is that if person A is in the park is equivalent to person B being in the park then person A must be person B)

Hence $v\in\{v-T(v)\mid v\in V\} $

Hence $\ker(T) \subseteq \{v-T(v)\mid v\in V\}$ as required.

Answer 1

Suppose $w \in \ker(T)$. In order to show that $w \in \{v-T(v) \mid v \in V\}$ as well, you need to produce a choice of $v \in V$ such that $w = v - T(v)$. There is a very simple choice of $v$ that will make this work.