Let $X$ be a normed linear space, and let $T$, $S$ be two different bounded linear operators on $X$ such that $T^2=T$, $S^2=S$, $TS=ST$. Show that $\lVert T-S\rVert \geq 1$.
This was a question on a prelim exam I took yesterday. In talking with some classmates afterward, it emerged that none of us figured it out.
I succeeded in showing that $\lVert T \rVert \geq 1$ and likewise for $S$, since: $$\lVert T^2 \rVert\leq \lVert T\rVert^2 \implies \lVert T \rVert\leq \lVert T\rVert^2 \implies \lVert T \rVert\geq 1$$ But this didn't seem to help me. After that, my strategy was to try a reductio, assuming $\lVert T-S\rVert <1$ and then multiplying both sides by either $\lVert T-S\rVert$ or $\lVert T+S\rVert$. None of this seemed to take me anywhere. Any tips would be appreciated.
Assume $\|T-S\|<1$. Let $Tv=0$ and set $w=Sv$. Then $$ S w = S^2 v = Sv = w \ \ \mbox{and} \ \ T w = T S v=STv =0$$ so $|w| = |(S-T)w|\leq \|S-T\| \; |w|$ implying $w=0$. Thus $S$ and $T$ have the same kernels. Similarly $1-S$ and $1-T$ have same kernels so $S$ and $T$ also have the same images. It follows that $S=T$.