Proving KerD^2=KerD and ImD=ImD^2 with Linear Transformations

ergonomics
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If i am given a linear transformation D:A->A,that is followed by
A=ImD(+)kerD
and i am asked to prove that kerD^2=kerD and imD=imD^2.

instead of trying to work it out the hard way by showing that every element of KerD is an element of kerD^2 , both directions.

would it not be easier to just say that dimA=dimA and hence the two structures are isomorphic which means that KerD={0} and ImD=A.

same goes for D^2:A->A
KerD^2={0}
ImD^2=A

=> therefore KerD^2=KerD and ImD^2=ImD ?
 
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ergonomics said:
would it not be easier to just say that dimA=dimA and hence the two structures are isomorphic which means that KerD={0} and ImD=A.
Why does it mean that?
 
would it not be easier to just say that dimA=dimA and hence the two structures are isomorphic
What two structures?
which means that KerD={0} and ImD=A.
Huh? I'm not sure what in the world you're doing, but just because A is isomorphic to A doesn't mean that every linear map D:A->A is an isomorphism. Is that what you were thinking?
 
According to my book two vector spaces of the same dimension are isomorphic to each other.
and the proof also apparently seems to be pretty simple.
If B is a basis for A, then we can easily show that ImD=A
and that T is injective and if T is injective then KerD={0}
 
If B is a basis for A, then we can easily show that ImD=A
and that T is injective and if T is injective then KerD={0}
You're forgetting one of the hypotheses for the theorem -- you need D to be an isomorphism.
 
yes akg unfortunately that is what i was thinking, that if the two were isomorphic to each other, then the map would necessairly be an isormophism.

just a few minutes before hurkyl put his post up, i was about to say that i went over the theorems in my book again, and that my line of thought was incorrect.

anyway, thank you all.
 
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