Answer to Question #213708 in Linear Algebra for anuj

Question #213708

Show that if S and T are linear

transformations on a finite dimensional

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


1
Expert's answer
2021-07-26T13:14:45-0400


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

According to question,

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

Then "S\\circ T :V \\rightarrow V" is a linear transformation.

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

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

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

and "dim(ker(ST))=nullity (ST)" .

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

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

Where  is a zero vector of "V."

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

Now , if "x\\in ker(T)" then

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

Hence, "x\\in Ker(ST)."

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

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

I,e,"nullity (T)\\leq 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)\\leq rank(T)" .



2. Consider the matrix ,

"S=\\begin{pmatrix}\n 1 & 0 \\\\\n 0 & 0\n\\end{pmatrix}" and "T=\\begin{pmatrix}\n 0 & 0 \\\\\n 1 & 0\n\\end{pmatrix}" and "ST=\\begin{pmatrix}\n 0 & 0 \\\\\n 0 & 0\n\\end{pmatrix}"

Then ,clearly "S,T:\\R^2\\rightarrow \\R^2" are two linear transformation.

Since , every "n\u00d7n" matrix from "\\R^n \\ to \\ \\R^n" is a linear transformation.

But "rank(ST)=0<rank(T)=1" .



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