Question #212365

Show that if S and T are linear

transformations on a finite dimensional

vector space, then rank (ST)<= rank (S).


1
Expert's answer
2021-07-01T11:51:45-0400

As VV is a finite dimensional vector space, let dimension of V is nV \ is \ n .

According to question,

Let , T:VVT:V\rightarrow V and S:VVS:V\rightarrow V are two linear transformation from a finite dimensional vector space VV to itself.

Then ST:VVS\circ T :V \rightarrow V is a linear transformation.

Now, dim(ImT)=rank(T) , dim(ImS)=rank(S)dim(ImT)=rank(T) \ ,\ dim(ImS)=rank(S)

and dim(Im(ST))=dim(Im(ST))=rank(ST).dim(Im(ST))=dim(Im(S\circ T))=rank (ST).

dim(Ker(S))=nullity(S),dim(ker(T))=nullity(T)dim(Ker (S))=nullity (S),dim(ker(T))=nullity(T)

and dim(ker(ST))=nullity(ST)dim(ker(ST))=nullity (ST) .

Note: Generally, ST\ S\circ T written as STST .

Now ,Ker(ST)={xV:(ST)(x)=0}Ker(ST)=\{x\in V :(ST)(x)=0 \} .

Where is a zero vector of V.V.

={xV:S(T(x))=0}=\{ x\in V:S(T(x))=0 \}

Now , if xker(T)x\in ker(T) then

T(x)=0     S(T(x))=S(0)=0T(x)=0 \ \implies S(T(x))=S(0)=0 .

Hence, xKer(ST).x\in Ker(ST).

Therefore,Ker(T)Ker(ST).Ker(T)\sube Ker(ST).

Thus,dim(Ker(T))dim(Ker(ST))dim(Ker(T))\leq dim(Ker(ST))

I,e,nullity(T)nullity(ST)...............(1)nullity (T)\leq nullity (ST)...............(1)

Now , from the rank-nullity theorem ,

rank(T)+nullity(T)=rank(ST)+nullity(ST)=nrank(T)+nullity(T)=rank(ST)+nullity (ST)=n

Now , from (1) we concluded ,

rank(ST)rank(S)rank(ST)\leq rank(S) .


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