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Maybe_Memorie
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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!