Question

Show that if S and T are linear

transformations on a finite dimensional

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

Answer 1

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 ,

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

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

Since , every $n×n$ matrix from $\R^n \ to \ \R^n$ is a linear transformation.

But $rank(ST)=0<rank(T)=1$ .