Show that $\mathrm{im}(T - I) \subseteq \mathrm{ker}(T+I)$ if $T^{2} = I$

Question

Suppose $T : V \rightarrow V$ is a linear map defined as $T^{2} = I$ where $I$ is the identity transformation.

I need to show that:

A) $\thinspace \thinspace \thinspace \mathrm{im}\left(T -I \right) \subseteq \mathrm{ker}\left(T + I \right)$

as well as

B) $\thinspace \thinspace \thinspace \mathrm{im}\left( T + I \right) \subseteq \mathrm{ker}\left(T - I \right)$.

I'm not sure what they mean by the addition and subtraction of two transformations. For A Do we start off by letting $v \in V$ and then applying $T(v)$ and $I(v)$ resulting in $T(v) - I(v) = T(v) -v$? And then we apply the transformation again: $T\left[T(v) -v \right] = T^{2}(v) - T(v) = I(v) - T(v) = v - T(v)$? I'm lost after that.

Answer 1

For part $(i)$ let $v\in \text{im}(T-I)$ Then, $v=(T-I)w$ for some $w\in V$. Now, $$(T+I)v=(T+I)\big((T-I)w\big)=(T+I)(Tw-w)$$$$=T^2w-Tw+Tw-w=T^2w-w=(T^2-I)w=0.$$ Therefore, $(T+I)v=0$. This implies $v\in\text{ker}(T+I)$.

For part $(ii)$ let $v\in \text{im}(T+I)$ Then, $v=(T+I)w$ for some $w\in V$. Now, $$(T-I)v=(T-I)\big((T+I)w\big)=(T-I)(Tw+w)$$$$=T^2w+Tw-Tw-w=T^2w-w=(T^2-I)w=0.$$ Therefore, $(T-I)v=0$. This implies $v\in\text{ker}(T-I)$.