Let $V$ be finite dimensional vector space. $T$ is a linear transformation from $V$ into $V$ and $E$ is a subspace of $V$.

Question

Shows, $T^{-1}(E)={\{u\in V|T(u)\in E}\}$

Prove dim$(T^{-1}(E))$=dim($ker(T)$)+dim($E\cap Im(T)$)

I know eventually I have to prove that dim$(T^{-1}(E))$= dim($ker(T^{-1}(E)$)+dim($Im(T^{-1}(E)$).

but I can't figur out the relationship between them.

Answer 1

Consider the restiriction $T|_{T^{-1}E}: T^{-1}(E) \to E$, then the ordinary dimension formula for morphism gives preciswely the desired formula, since $Im(T|_{T^{-1}E})=E \cap Im(T) $

Answer 2

Demo.

Proof. $\blacktriangleleft$ First we verify that $\mathcal T^{-1}E$ is a vector space. If $\alpha, \beta \in \mathcal T^{-1}(E)$, then $\mathcal T \alpha + \mathcal T\beta = \mathcal T (\alpha + \beta) \in E$, thus $\alpha + \beta \in \mathcal T ^{-1}(E)$. Also we could see that $k\alpha \in \mathcal T^{-1 }(E)$ for $\alpha \in \mathcal T^{-1}(E)$ and $k \in \mathbb F$, where $\mathbb F $ is the base field. Note that $E$ is a subspace of $V$, then we have $E \supset \{0\}$. Thus $\mathcal T^{-1} (E) \supset \mathcal T^{-1}(\{0\}) = \mathrm {Ker}\, \mathcal T$. Therefore we could deduce that $$ \mathrm {Ker}\left(\mathcal T|_{\mathcal T^{-1}(E)} \right) = \mathrm {Ker}\, \mathcal T. $$ Now apply the Rank-Nullity theorem to the surjection $\mathcal S:= \mathcal T |_{\mathcal T^{-1}(E)} \colon \mathcal T^{-1}E \to E \cap \mathrm {Im} \, \mathcal T$, then $$ \dim (\mathcal T^{-1}E) = \dim (\mathrm {Ker}\,\mathcal S) + \dim (\mathrm {Im}\, \mathcal S) = \dim (\mathrm {Ker}\, \mathcal T) + \dim(E \cap \mathrm {Im}\,\mathcal T), $$ and this is the equation we wanted. $\blacktriangleright$