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

  1. #1
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    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
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  2. #2
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    Rank + Nullity = n
    So if Rank = Nullity then it means n must be even and:
    Rank = Nulllity = n/2.
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  3. #3
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    help on the reverse

    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.
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  4. #4
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    Quote Originally Posted by sunlight View Post
    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 n. From the rank+nullity theorem you know that kerT also has dimension n. But if two spaces have the same dimension and one is a subspace of the other, then they must coincide.
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  5. #5
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    Thumbs up Thank you

    Quote Originally Posted by Opalg View Post
    The rank of T is the dimension of ImT. So ImT has dimension n. From the rank+nullity theorem you know that kerT also has dimension n. 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.
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