Let \(V,W\) be finite-dimensional vector spaces, let \(X\subseteq W\) be a subspace, and let \(T:V\to W\) be linear. Prove that \[ \dim T^{-1}(X)\ge \dim V-\dim W+\dim X. \]
Solution
1. The restriction of \(T\) to \(T^{-1}(X)\) has image\(X\cap\operatorname{im}T\) and kernel \(\ker T\).
If \(v\in T^{-1}(X)\), then \(T(v)\in X\) and, by definition, \(T(v)\) lies in \(\operatorname{im}T\). Thus the image of the restriction is contained in \(X\cap\operatorname{im}T\). Conversely, if \(w\in X\cap\operatorname{im}T\), then \(w=T(v)\) for some \(v\in V\), and \(w\in X\) implies \(v\in T^{-1}(X)\). Hence \(w\) lies in the image of the restriction. Its kernel consists exactly of those \(v\) with \(T(v)=0\), namely \(\ker T\).
2.\[ \dim T^{-1}(X) = \dim\ker T+\dim(X\cap\operatorname{im}T). \]
Apply the rank-nullity theorem to the restricted map described in step 1.
3.\[ \dim(X\cap\operatorname{im}T) \geq \dim X+\dim\operatorname{im}T-\dim W. \]
The dimension formula for the subspaces \(X\) and \(\operatorname{im}T\) gives \[ \dim(X\cap\operatorname{im}T) = \dim X+\dim\operatorname{im}T -\dim(X+\operatorname{im}T). \] Since \(X+\operatorname{im}T\subseteq W\), \[ \dim(X+\operatorname{im}T)\leq\dim W, \] which yields the claimed inequality.
4.\[ \dim T^{-1}(X) \geq \dim V-\dim W+\dim X. \]
Combining steps 2 and 3 gives \[ \dim T^{-1}(X) \geq \dim\ker T+\dim X+\dim\operatorname{im}T-\dim W. \] By rank-nullity for \(T:V\to W\), \[ \dim\ker T+\dim\operatorname{im}T=\dim V. \] Substitution proves the desired inequality.
5. Q.E.D.
Step 4 is the required conclusion.