Let $V$ be a finite dimensional $K$-vector space of dimension $n$, and $T$ a linear transformation. I want to find a bijection between $T$-invariant subspaces of dimension $m$ and $T$-invariant subspaces of dimension $n-m$.
** The profesor told us to use the dual transformation $T^{*}$. I proved that if $W$ is $T$-invariant, then $W^{0}$ (the annihilator of $W$) is $T^{*}$-invariant.
You have proved that if $W$ is $T$-invariant, then $W^0$ is $T^*$-invariant. That defines a bijection from the dimension $m$ subspaces of $V$ and the dimension $n-m$ subspaces of $T^*$.
Let $\Phi$ denote the map from the set of dimension $m$ $T$-invariant subspaces to the set of dimension $n-m$ $T^*$-invariant subspaces defined by $\Phi(U) = U^0$. Then, $\dim(U^0) = n - \dim(U) = n-m$, and your statement shows that the range of $\Phi$ lies within the set of $T^*$-invariant subspaces. In order to show that $\Phi$ is bijective, it suffices to note that the map $\Psi(U) = \iota(U^0)$, where $\iota : V^{**} \to V$ denotes the canonical identification of $V^{**}$ with $V$, is the inverse of $\Phi$.
Now, we'll define a bijection between the dimension $n-m$ subspaces of $T^*$ and the dimension $n-m$ subspaces of $T$. To that end, we note that $T$ is always similar to its dual map $T^*$ (as can be proved with the help of rational canonoical form AKA Frobenius normal form). That is, there exists a linear map $P:V \to V^*$ such that $$ T^* = P \circ T \circ P^{-1}. $$ Given such a map $P$, we can define our bijection by $\mu_P(U) = P(U)$.
With that, $\mu_P \circ \Phi$ defines the desired bijection.