Rank/Image/Kernel proof

  • #1

Homework Statement



Let V be a vector space. Show that for every three linear operators
A,B,C: V -> V, we have
rk(ABC) =< rk(B)

Homework Equations



V = rk(A) + dimKer(A)

rk(A) = dimIm(A)

The Attempt at a Solution



Im(ABC) = {ABC(v) | vEV}
= {AB(C(v)) | vEV}

So Im(ABC) is a subset of Im(AB)

So dimIm(ABC) =< dimIm(AB)

So rk(ABC) =< rk(AB)

Im(AB) = {A(B(w)) | wEV}

Ker(B) subset of ker(AB) because if Bx=0, then ABx = A0 = 0

By the dimension formula, this leads to rk(B) >= rk(AB)

So putting the two results together we get rk(ABC) =< rk(B)


Is this correct? Thanks!
 

Answers and Replies

  • #2
22,129
3,297
Hi Maybe_Memorie! :smile:

Your proof seems correct! But it seems you'll need V to be finite dimensional for it to work.
 
  • #3
Hi Maybe_Memorie! :smile:

Your proof seems correct! But it seems you'll need V to be finite dimensional for it to work.

We've only dealt with finite dimensional vector spaces.
 

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