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

[ergonomics]

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 ?

 

[Hurkyl]

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?

 

[AKG]

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?

 

[ergonomics]

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}

 

[Hurkyl]

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.

 

[ergonomics]

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.