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## Main Question or Discussion Point

Problem:

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.

Attempt:

If I call:

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!

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.

Attempt:

If I call:

- a: 0 → ker(f),
- b: ker(f) → V,
- c: V → im(f),
- d: im(f) → 0.

- 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!