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I Prove the sequence is exact: 0 → ker(f) → V → im(f) → 0

  1. Dec 4, 2016 #1
    Let f ∶ V → V be a linear operator on a finite-dimensional vector space V .
    Prove that the sequence 0 → ker(f) → V → im(f) → 0 is exact at each term.

    If I call:
    • a: 0 → ker(f),
    • b: ker(f) → V,
    • c: V → im(f),
    • d: im(f) → 0.
    Then the sequence is exact at:
    • ker(f) if im(a)=ker(b),
    • V if im(b)=ker(c),
    • im(f) if im(c)=ker(d).

    I can show this for the following:

    At im(f):
    im(c)={c(v) | v∈V }=im(f)={e∈im(f) | d(e)=0}=ker(d)

    At V:
    im(b)={b(v) | v∈ker(f)}=ker(f)={e∈V | c(e)=0 ∈ im(f)}=ker(c)

    I'm having trouble showing it is exact at ker(f) though. I know that im(a)=0 and ker(b)={v∈ker(f) | b(v)=0 ∈V} however doesn't this mean that ker(b)=ker(f), and so how can I deduce that ker(b)=ker(f)=im(a)=0?

    Thanks for any help!
  2. jcsd
  3. Dec 4, 2016 #2


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    Staff: Mentor

    You must have a look on the definition of ##b : \ker(f) \rightarrow V##. E.g. if we defined ##b =0## then ##b## would still be linear and a map from ##\ker (f)## to ##V##. So the way how ## b## is defined is crucial here.
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