A result involving dimension of inverse image (Vector spaces) Let $V$ be a vector space, $T:V \to V$ a linear operator and consider $W \subset V$ a subspace. I was thinking about the follow inequality:
$$\dim T^{-1}(W) \leq \dim \ker T + \dim W.$$
How can I prove that? I've tried this way:
$$\dim W + \dim \ker T = \dim W + \ker T - \dim W \cap \ker T \geq \dim W+ \ker T.$$ Then if $T^{-1}(W) \subset W + \ker T$, the result follows. But I guess it is not correct.
Can you help me?
 A: Clearly, $\ker T\subset T^{-1}(W)$. Let $\{e_1,\ldots,e_k\}$ be a basis of $\ker T$ and extend it to a basis $\{e_1,\ldots,e_n\}$ of $T^{-1}(W)$. Then $\bigl\{T(e_{k+1}),\ldots,T(e_n)\bigr\}$ is a linearly independent subset of $W$, because if there were scalars $a_{k+1},\ldots,a_n$, not all of which are equal to $0$, such that $a_{k+1}T(e_{k+1})+\cdots+a_nT(e_n)=0$, then $T(a_{k+1}e_{k+1}+\cdots+a_ne_n)=0$, which means that $a_{k+1}e_{k+1}+\cdots+a_ne_n\in\ker T$, which is impossible, by the way that $e_{k+1},\ldots,e_n$ were chosen. Since $\bigl\{T(e_{k+1}),\ldots,T(e_n)\bigr\}$ is a linearly independent subset of $W$ with $n-k$ elements, $n-k\leqslant\dim W$.
Therefore\begin{align}n&=\dim T^{-1}(W)\\&=k+(n-k)\\&\leqslant\dim\ker T+\dim W.\end{align}
A: If $T|_{T^{-1}(W)} : T^{-1}(W) \to V$ is the restriction of $T$ on $T^{-1}(W)$, we see that
\begin{align} \ker T|_{T^{-1}(W)} &= (\ker T) \cap T^{-1}(W) \\ &= T^{-1}(\{0\}) \cap T^{-1}(W) \\ &= T^{-1}(\{0\} \cap W) \\ &= T^{-1}(\{0\}) = \ker T \end{align}
and that $$\operatorname{im} T|_{T^{-1}(W)} = T(T^{-1}(W)) = W \cap \operatorname{im} T.$$ Thus, by the rank-nullity theorem we have that $$\begin{align} \dim T^{-1}(W) &= \dim \ker T|_{T^{-1}(W)} + \dim \operatorname{im} T|_{T^{-1}(W)} \\ &= \dim \ker T + \dim(W \cap \operatorname{im} T).\end{align}$$
