Question

Show that
Rank(ST)=Rank s ,if T is non singular
Where S,T : V->V are linear transformation of a finite dimensional vector space.

Answer 1

Let $S$ is represented by matrix $A(m\times n)$ and $T$ is represented by matrix $B(n\times n)$ (since $T$ is non singular). Then $S\circ T$ is represented by matrix $AB$.

From properties of rank: if $B$ is matrix of rank $n$, then

$rank(AB)=rank(A)$

Proof:

let C=AB, then

$C_{ij}=\Sigma A_{ik}B_{kj}\implies rank(C)\leq rank(A)$

$A=CB^{-1}$

if $det(B)\neq0$ , then $rank(C)=rank(A)$

In our case: $B$ is matrix of rank $n$ (since $det(B)\neq0$ )

So, we get: $rank(S\circ T)=rank(S)$