Thread: linear algebra problem: consequences of im(T)=ker(T) ?

I can't work out how to do this proof.

V is a vector space of dimension n where n >/= 1.
T: V --> V is a linear transformation.
You need to show that the following two are equivalent:

(a) im(T)=ker(T)

(b) T^2 = 0 AND n is even AND rk(T)=n/2

I think once you have proved that (a) implies (b), the other way round would be easy, and i've already shown that (a) implies T^2=0, but could anyone help me on how to show the other two parts of (b)?


Thanks very much

Rank + Nullity = n
So if Rank = Nullity then it means n must be even and:
Rank = Nulllity = n/2.

Please, help with the last 1/4 of the problem:

Prove that if T:V--->V is linear and dimV=n>1 then (a) ImT=kerT and (b) T^2=0, rkT= 1/2.n and n - even are equivalent.

I have shown that (a) ---> (b) and that if (b) then ImT is a subset of kerT. Somehow I got stuck on "KerT is a subset of ImT".

Any help is deeply appreciated.

Originally Posted by sunlight

Please, help with the last 1/4 of the problem:

Prove that if T:V--->V is linear and dimV=n>1 then (a) ImT=kerT and (b) T^2=0, rkT= 1/2.n and n - even are equivalent.

I have shown that (a) ---> (b) and that if (b) then ImT is a subset of kerT. Somehow I got stuck on "KerT is a subset of ImT".

Any help is deeply appreciated.

The rank of T is the dimension of ImT. So ImT has dimension ½n. From the rank+nullity theorem you know that kerT also has dimension ½n. But if two spaces have the same dimension and one is a subspace of the other, then they must coincide.

Originally Posted by Opalg

The rank of T is the dimension of ImT. So ImT has dimension ½n. From the rank+nullity theorem you know that kerT also has dimension ½n. But if two spaces have the same dimension and one is a subspace of the other, then they must coincide.

Thank you very much indeed. I used this helpful observation (two spaces with same dimension if one subset of other then they coincide) in another problem too.