 Solution

The preimage isIt contains $0$, and if $u,v$ belong to it and $a,b\in \mathbb{R}$, then $T(au+bv)=aT(u)+bT(v)\in X$, since $X$ is a vector subspace. Thus $T^{-1}(X)$ is a vector subspace. It contains the kernel of $T$ because $0\in X$.

To establish the dimension of the preimage of a vector subspace formula without assuming the theorem to be deduced, choose a basis $k_1,\dots ,k_{n_T}$ of $\ker T$ and a basis $y_1,\dots ,y_d$ of $X\cap \mathrm{im}T$, where $d=\dim (X\cap \mathrm{im}T)$. For each $j$, choose $u_j\in U$ with $T(u_j)=y_j$. These $u_j$ lie in $T^{-1}(X)$.

The family $k_1,\dots ,k_{n_T},u_1,\dots ,u_d$ is linearly independent: applying $T$ to a vanishing linear combination forces all $u_j$ coefficients to vanish by linear independence of the $y_j$, and then all remaining coefficients vanish by linear independence of the $k_i$. It also spans $T^{-1}(X)$: for $u$ in that preimage, write $T(u)={\sum }_j{b}_j{y}_j$. Then $u-{\sum }_j{b}_j{u}_j\in \ker T$, so it is a linear combination of the $k_i$. ConsequentlySet $X=V$. Then $T^{-1}(V)=U$ and $V\cap \mathrm{im}T=\mathrm{im}T$, giving the rank-nullity theorem: