Q:

Prove that a linear map $A$ does not have an inverse map if $A \circ A$ = O where $A$ is a linear transformation from X onto itself

My solution is this: Since $A$ maps from a finite dimension vector space onto the same space, if $A$ is injective it should also be surjective.  If $A$ is injective then $A(A(x))$ should also be injective which is a contradiction since $A(A(x))$ is equal to the zero $(O)$ map.  Hence $A$ is not injective implying it isn't surjective implying an inverse mapping cannot exist.  Is this good enough or should I be rigorous in some other way? 

A:

Violation in injectivity is apparent, which is sufficient proof in my opinion.  Not that apparent, however, how you deduce that lack of injectivity implies it's not surjective.   
Of course it should be shown rigorously that $A$ is not injective and that concludes the proof, because invertibility requires bijectivity.  If there is no injectivity, then there is no bijectivity.