Let $T : U → V$ and $S : V → W$ be linear transformations. How is $\operatorname{rank}(ST)$ related to $\operatorname{rank}(T)$?

Question

Let $T : U → V$ and $S : V → W$ be linear transformations. How is $\operatorname{rank}(ST)$ related to $\operatorname{rank}(T)$? I know that $\operatorname{img}(S \circ T) = \operatorname{img}(S)$. I believe this implies that $\operatorname{rank}(ST) = \operatorname{rank}(S)$ but I believe there is some relation to $\operatorname{rank}(T)$

Since $\operatorname{rank}(S) = \dim(V) - \operatorname{null}(S)$, I believe finding a relationship between $\operatorname{null}(S)$ and $\operatorname{rank}(T)$ would help me here, but I don't think I can make any assumptions about how they are related since the kernel and image and their bases could change depending on the $T$ and $S$ I choose. Any help hear would be appreciated.

Answer 1

The only statement you can do without further assumptions is that $$ \operatorname{rank}(ST)\le\operatorname{rank}(T) $$ You can also say that $\operatorname{rank}(ST)\le\operatorname{rank}(S)$, just because the image of $ST$ is a subspace of the image of $S$.

It is generally false that $\operatorname{im}(ST)=\operatorname{im}(S)$, because $T$ might map something in the kernel of $S$, even everything if it is the zero map.

How can you prove the first statement? With the rank nullity theorem. Indeed you have \begin{align} \dim U &=\operatorname{rank}(T)+\operatorname{null}(T)\\ &=\operatorname{rank}(ST)+\operatorname{null}(ST) \end{align} and it is immediate to see that the kernel of $T$ is a subspace of the kernel of $ST$, forcing $\operatorname{null}(T)\le\operatorname{null}(ST)$, whence the statement follows from $$ \operatorname{rank}(T)-\operatorname{rank}(ST)=\operatorname{null}(ST)-\operatorname{null}(T) $$ It's easy to make examples where any rank less than or equal to $\operatorname{rank}(T)$ is attained.

Take a basis $\{v_1,v_2,\dots,v_k\}$ of $\operatorname{im}(T)$ and complete it to a basis $\{v_1,\dots,v_k,v_{k+1},\dots,v_n\}$ of $V$. Then you can define, for $0\le m\le k$, the linear map $S_m\colon V\to V$ (so $W=V$) with $$ S_m(v_i)=\begin{cases} v_i & 1\le i\le m \\[6px] 0 & m<i\le n \end{cases} $$ and $\operatorname{rank}(S_mT)=m$.

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Prompted by a comment, here's an addition. Restrict $S$ to a linear map $S'\colon\operatorname{im}(T)\to W$. Then $\ker(S')=\operatorname{im}(T)\cap\ker(S)$ and $\operatorname{im}(ST)=\operatorname{im}(S')$, so the rank-nullity theorem tells you that $$ \dim\operatorname{im}(T)=\operatorname{rank}(T)=\operatorname{rank}(ST)+\dim(\operatorname{im}(T)\cap\ker(S)) $$ Using the previous equalities, we can state that $$ \operatorname{null}(ST)-\operatorname{null}(T)=\dim(\operatorname{im}(T)\cap\ker(S)) $$

Answer 2

Firstly, the statement $im ~(S\circ T)=im~ S$ is not true, as suggested in the comment. Yes, if you impose some conditions on $T$, then you may connect $rank~(S\circ T)$ with rank $T$. One of such instance is, when $T$ is Bijective.

If $T$ is bijective, then $S(x)=S(T (T^{-1}(x)))$ for arbitrary $X$, which means $im ~(S\circ T)=im~ S$. Then $rank~(S\circ T)=rank~S.$

Thus, you can say, when $T$ (assuming a square matrix identified as an operator) is of full rank then $rank~S$ remains unchanged after multiplied(composed) by $T$.