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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!