Let $T : V → V$ be a linear transformation. Show that $T^2 = I $ if and only if $V = ker(T − I) ⊕ ker(T + I)$
answer:
let $U=(T-I)$ , and $W=(T+I)$
$V=U \bigcap W$
$V\in ker(T-I) : (T-I)(v)=0, → Tv-v=0$
$V\in ker(T+I) : (T+I)(v)=0, → Tv+v=0$
$ [Tv-v=0]-[Tv-v=0] → v=0$
$V=V+Tv-Tv$
Am i on the right way to solve the question and how to complete it, and if not would someone help me through !!!
Since for any $v\in V$ $$ v=\frac1{2}((T+I)v-(T-I)v) $$ We have $$V=W+U$$ where $$ x=\frac1{2}(T+I)v, \:x\in W\quad\text{and}\quad y=\frac1{2}(I-T)v, \:y\in U $$ If there is a $z\in W\cap U$, then $$ z=\frac1{2}(T+I)v'=-\frac1{2}(T-I)v' $$ Hence we have $v'=0$ and so $z=0$. Thus $W\cap U=0$ and we conclude $$V=W\oplus U$$
Since $$ (T-I)x=\frac1{2}(T-I)(T+I)v=0 $$ There is $$ W=\ker{(T-I)} $$ Likewise since $$ (T+I)y=-\frac1{2}(T-I)(T+I)v=0 $$ We have $$ U=\ker{(T+I)} $$ Hence $$ V=\ker{(T-I)}\oplus \ker{(T+I)} $$
Assume first that $T^2 =I$. Then, we note automatically that $\ker (T-I) \cap \ker (T+I) = 0$, since if $v \in \ker(T-I)\cap \ker (T+I)$, then we see that $v = -v$, its additive inverse, so that $v = 0$ identically.
Now we employ rank nullity theorem: We know that $\dim V = \dim \textrm{im} (T-I) + \dim \ker (T-I)$, so that if we choose $v \notin \ker (T-I)$, we conclude that $v \in \textrm{im} (T-I)$ (basically, $V = \ker (T-I) + \textrm{im} (T-I)$).
Using this, assume this is so. Then, $v = (T-I)(u)$ for some $u \in V$. Now apply $T+I$ to both sides: We have: $$(T+I)(v) = (T+I)(T-I)(u) = (T^2 - I)(u) = 0$$
Since $T^2 = I$. Thus, $\ker (T+I) = \textrm{im} (T-I)$. We can put this together with the above to conclude $V = \ker (T-I) \oplus \ker (T+I)$.
Conversely, suppose $V = \ker (T-I) \oplus \ker (T+I)$. Then, given $v \in V$, we know it is of the form $v = u + w$ with $u \in \ker (T-I)$, $w \in \ker (T+I)$.
Then apply $T$ once to both sides:
$$T(v) = T(u) + T(w) = u - w$$
Apply $T$ once more:
$$T^2 (v) = T(u) - T(w) = u + w = v$$
So that $T^2 = I$, as desired.
