MHB Linear Map Problem: Proving Equivalence of a & b

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The discussion centers on proving the equivalence of two statements regarding a linear map T in a vector space V. The first statement asserts that the kernel of T is trivial (Ker T = {0}), while the second states that if T squared applied to any vector v equals zero, then T applied to that vector must also equal zero. The user outlines a proof for one direction, showing that if T^2(v) = 0, then T(v) must equal zero. They seek assistance in proving the reverse direction, demonstrating that if the intersection of the image and kernel of T is trivial, then T^2(v) = 0 implies T(v) = 0. The discussion highlights the logical connections between these properties of linear maps.
Fernando Revilla
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I quote an unsolved problem from MHF (Linear map problem) posted by user jdm900712

et V be a vector space over the field F. and T
png.latex
L(V, V) be a linear map.

Show that the following are equivalent:
a) I am T
png.latex
Ker T = {0}
b) If T^2(v) = 0 -> T(v) = 0, v
png.latex
V
Using p -> (q -> r) <-> (p
png.latex
q) ->r
I suppose I am T
png.latex
Ker T = {0} and T
png.latex
(v) = 0.
then I know that T(v)
png.latex
Ker T and T(v)
png.latex
I am T
so T(v) = 0.
I need help on how to prove the other direction.
http://mathhelpforum.com/advanced-algebra/212908-linear-map-problem.html
 
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We have to prove $\mbox{Im }T\cap \ker T=\{0\}\Leftrightarrow (T^2(v)=0\Rightarrow T(v)=0)$

$\Rightarrow)$ Suppose $T^2(v)=0$, then $T(T(v))=0$. But $T(v)\in \mbox{Im }T$ (by definition of image) and $T(v)\in \ker T$. By hypothesis, $T(v)=0$.

$\Leftarrow)$ Suppose $x\in\mbox{Im }T\cap \ker T$ then, $x\in\mbox{Im }T$ and $x\in \ker T$, that is $x$ has de form $x=T(w)$ and $T(x)=0$. This implies $T(x)=T^2(w)=0$. By hypothesis $T(w)=x=0$, so $\mbox{Im }T\cap \ker T=\{0\}$.
 
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