I was posed the following question : If T is a linear operator on vector space V and given that the kernel and range of T are disjoint, ie. they have only the zero vector in their intersection and T composed T, ie. T^2(v)=0, for each vector v in V, I need to show that T(v)=0.
I made several attempts, but I haven't found anything substantial.
Any help is appreciated. But hints are appreciated more than the answer itself. I'd like to try myself.
Best Answer
Take the equation $T^2(\vec v)=\vec0$ and rewrite it as $T(T(\vec v))=\vec0$. Now let $\vec u= T(\vec v)$. Your rewritten equation now implies that $T(\vec u)=\vec 0$.
Since you asked for hints, I'll let you take it from here.