Finding Dimension of Preimage of Subspace under Surjective Linear Transformation: A Linear Algebra Problem

Question

"I am currently working on solving a previous year Linear Algebra exam question. The question involves vector spaces $V_1$ and $V_2$ over the same field $F$, with a surjective linear transformation $T: V_1 \rightarrow V_2$. Given that $W$ is a subspace of $V_2$ with dimension 2, and $dim V_1 = 7$ and $dim V_2 = 5$, the task is to determine $dim T^{-1}(W)$.

My Approach: I plan to start by examining the restriction of the linear transformation $T_1: T^{-1}(W) \rightarrow V_2$. Since $T$ is onto, we know that $T(T^{-1}(W)) = W$. To proceed, I intend to apply the rank-nullity theorem for $T_1$, which states that $dim(null(T_1)) + dim(Im(T_1)) = dim(T^{-1}(W))$. However, I'm currently unsure how to calculate $dim(null(T_1))$ and $dim(Im(T_1))$. Any guidance on this problem would be greatly appreciated. Thank you in advance."

Answer 1

HINT: being surjective means that $\dim(\operatorname{null}T)=2$ as $\dim(\operatorname{img}T)=\dim V_2=5$. Now, as $T$ is surjective, the map $\tilde T:V_1/\operatorname{null}T \to V_2$ given by $\tilde T(\operatorname{v+null}T)=Tv$ is an isomorphism.

Thus $H:=\tilde T^{-1}(W)$ have dimension $2$ and observe that $T=\pi \circ \tilde T$ where $\pi:V_1\to V_1/(\operatorname{null}T),\, v\mapsto v+\operatorname{null}T$ is the canonical projection, therefore $T^{-1}(W)=\pi^{-1}(H)$. Can you follow from here?

As $\dim H=2$ there are $a,b\in V_1$ such that $H=\operatorname{span}\{a+\operatorname{null}T,b+\operatorname{null}T\}$. Now, as $\dim (\operatorname{null}T)=2$ also, there are $c,d\in V_1$ such that $\operatorname{null}T=\operatorname{span}\{c,d\}$. Thus $\pi^{-1}(H)=\operatorname{span}\{a,b,c,d\}$ so $\dim T^{-1}(W)=4$.∎