As V is a finite dimensional vector space, let dimension of V is n .
According to question,
Let , T:V→V and S:V→V are two linear transformation from a finite dimensional vector space V to itself.
Then S∘T:V→V is a linear transformation.
Now, dim(ImT)=rank(T) , dim(ImS)=rank(S)
and dim(Im(ST))=dim(Im(S∘T))=rank(ST).
dim(Ker(S))=nullity(S),dim(ker(T))=nullity(T)
and dim(ker(ST))=nullity(ST) .
Note: Generally, S∘T written as ST .
Now ,Ker(ST)={x∈V:(ST)(x)=0} .
Where is a zero vector of V.
={x∈V:S(T(x))=0}
Now , if x∈ker(T) then
T(x)=0 ⟹S(T(x))=S(0)=0 .
Hence, x∈Ker(ST).
Therefore,Ker(T)⊆Ker(ST).
Thus,dim(Ker(T))≤dim(Ker(ST))
I,e,nullity(T)≤nullity(ST)...............(1)
Now , from the rank-nullity theorem ,
rank(T)+nullity(T)=rank(ST)+nullity(ST)=n
Now , from (1) we concluded ,
rank(ST)≤rank(S) .
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