$T:U \to V$, $S: V\to W$, $P:W\to U$, Range($ST$)=nullspace($P$), nullspace($ST$)=range($P$), and rank($T$)=rank($S$)?

Question

$Range($ST$)=nullspace($P$)$ $\rightarrow rank(ST)=dim(W)-rank(P)$ and

$nullspace(ST) = range(P) \rightarrow dim(U)-rank(ST)=rank(P)$

Combining above two, $dim(U)=dim(W)$

Clearly, option $(A)$ can't be always true since $dim(V)$ is not necessarily equal to $dim(U)$

$(b)$ is not correct.

I am unable to rule out between $(c)$ and $(d)$

Answer 1

You can find a counter example for (d) using slight variations of the identity. Suppose $U$ and $W$ have dimension 3, and $V$ has dimension 4. Then if $T$ 1-to-1, rank($T$) = 3. If rank($T$) = rank($S$), then rank($S) = 3$, so dim(NS($S$)) = 1. Set up NS($S$) in the right way and you'll have your answer.