some confusion in Rank nullity theorem.

Let $U,V$ and $W$ be finite dimensional real vector spaces, $T: U→V,\;S:V→W$ and $P:W→U$ be linear transformations. If $\operatorname{range}(ST) = \operatorname{null space} (P)$, $\;\operatorname{null space}(ST)=\operatorname{range} P\;$ and $\operatorname{rank}(T)=\operatorname{rank}(S)$. Then which of the following is true?

Option 1) nullity of T= nullity of S

option 2) $\dim U \neq \dim W$

option 3) if $\dim V= 3$, $\;\dim U=4$ ,then $P$ is not identically zero

option 4) If $\dim V= 4$, $\;\dim U= 3$ and $T$ is one-one ,then $P$ is identically zero.

My thinking : By rank nullity theorem, I got only option 1 is correct as I don't know the other options...

Pliz help me and I'm very confused as any hints or solution can be appreciated.

Thank u

• @ Bernard help me...which option is true ?
– user525416
Mar 10 '18 at 15:44
• I'd love to, but it's very intricate. Actually, I don't see how you derive option 1). All I can say is that option 2) is false. I didn't examine the last two options yet. Mar 10 '18 at 15:51
• Option 2) is false. On the contrary, $\dim U=\dim W$, since $$\dim U=\dim (\ker ST)+\dim(\operatorname{Im}ST)=\dim(\operatorname{Im}P)+\dim (\ker P)=\dim W.$$
• Option 3) is true: if $\dim U>\dim V$, $T$ cannot be injective, nor $ST$. Hence $$\operatorname{Im}P = \ker ST\ne\{0\}.$$