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

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.

• If $v \in \text{Im}(T - I)$ then $v = T(x) - x$ for some $x\in V$. Hence, $$(T + I)(v) = T(v) + v = T(T(x) - x) + T(x) - x = 0,$$ which shows that $v \in \ker (T + I)$ Commented Nov 3, 2019 at 2:54

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 $$v\in\text{ker}(T-I)$$.