
Linear map problem
Let V be a vector space over the field F. and T $\displaystyle \in$ L(V, V) be a linear map.
Show that the following are equivalent:
a) Im T $\displaystyle \cap$ Ker T = {0}
b) If T^2(v) = 0 > T(v) = 0, v$\displaystyle \in$ V
Using p > (q > r) <> (p$\displaystyle \wedge$q) >r
I suppose Im T $\displaystyle \cap$ Ker T = {0} and T$\displaystyle ^{2}$(v) = 0.
then I know that T(v)$\displaystyle \in$ Ker T and T(v)$\displaystyle \in$ Im T
so T(v) = 0.
I need help on how to prove the other direction.

Re: Linear map problem
Hey jdm900712.
Can you make use of the ranknullity theorem for your map?

Re: Linear map problem
what you've done is shown a) implies b).
now suppose we have b). so T^{2}(v) = 0 implies v = 0.
suppose that w is in im(T) and ker(T). since w is in ker(T), T(w) = 0. since w is in im(T), w = T(v), for some vector v.
hence 0 = T(w) = T(T(v)) = T^{2}(v).
by b), this means that v must be 0. hence w = T(v) = T(0) = 0, so every element of the intersection of ker(T) and im(T) is a 0vector, that is: ker(T)∩im(T) = {0}, which is precisely a).