Let $\mathbb{F}[X]$ be the polynomial ring with one variable.
$V$ can be regarded as an $\mathbb{F}[X]$-module by defining $Xv = T(v)$ for every $v \in V$. $\mathbb{F}[X]$-submodules of $V$ are none other than $T$-invariant subspaces of $V$.
Let $K = \mathbb{F}[X]/(p_t)$.
Since $p_t$ is irreducible, $K$ is a field.
Since $p_t V = 0$, $V$ can be regarded as a $K$-module.
Let $W$ be a $T$-invariant subspace of $V$.
$W$ can be regarded as a $\mathbb{F}[X]$-submodule of $V$.
Since $p_t W = 0$, $W$ can be regarded as a $K$-submodule of $V$.
Hence there exists a $K$-submodule $W'$ such that $V = W \oplus W'$.
Since $W'$ is a $\mathbb{F}[X]$-submodule, it is $T$-invariant.
This completes the proof.
Let $\beta=\{w_1,w_2,\ldots,w_k\}$ and $\gamma=\{x_1,x_2,\ldots,x_m\}$ be the bases for $W$ and $W^\perp$, respectively. It suffices to show that
$$\beta\cup\gamma=\{w_1,w_2,\ldots,w_k,x_1,x_2,\ldots,x_m\}$$
is a basis for $V$.
Given $v\in V$, then it is well-known that $v=v_1+v_2$ for some $v_1\in W$ and $v_2\in W^\perp$. Also because $\beta$ and $\gamma$ are bases for $W$ and $W^\perp$, respectively, there exist scalars
$a_1,a_2,\ldots,a_k,b_1,b_2,\ldots,b_m$ such that
$v_1=\displaystyle\sum_{i=1}^ka_iw_i$ and $v_2=\displaystyle\sum_{j=1}^mb_jx_j$. Therefore
$$v=v_1+v_2=\sum_{i=1}^ka_iw_i+\sum_{j=1}^mb_jx_j,$$
which follows that $\beta\cup\gamma$ generates $V$. Next, we show that
$\beta\cup\gamma$ is linearly independent. Given
$c_1,c_2,\ldots,c_k,d_1,d_2,\ldots,d_m$ such that
$\displaystyle\sum_{i=1}^kc_iw_i+\sum_{j=1}^md_jx_j={\it 0}$, then
$\displaystyle\sum_{i=1}^kc_iw_i=-\sum_{j=1}^md_jx_j$. It follows that
$$\sum_{i=1}^kc_iw_i\in W\cap W^\perp\quad\mbox{and}\quad
\sum_{j=1}^md_jx_j\in W\cap W^\perp.$$
But since $W\cap W^\perp=\{{\it 0}\,\}$ (gievn $x\in W\cap W^\perp$,
we have $\langle x,x\rangle=0$ and thus $x={\it 0}\,$), we have
$\displaystyle\sum_{i=1}^kc_iw_i=\sum_{j=1}^md_jx_j={\it 0}$. Therefore
$c_i=0$ and $d_j=0$ for each $i,j$ becasue $\beta$ and $\gamma$ are bases
for $W$ and $W^\perp$, respectively. Hence we conclude that $\beta\cup\gamma$ is linearly independent.
Best Answer
No idea what $A(V)$ means, but if $U$ is $T$-invariant (i.e. $T(u) \in U$ for all $u \in U$), then $U^{\top} = \{ w \in V \, | \, \forall u \in U, \quad (w,u) = 0 \}$ will also be $T$-invariant.
Since $T$ is unitary, there exists $S$ such that $T \circ S$ is the identity map on $V$, so that $T$ is an isomorphism of $V$ to itself. By $T$-invariance, the restriction of $T$ to $U$ is an isomorphism of $U$ to itself, because $T$ maps $U$ to $U$ and is injective, but being a linear map of finite-dimensional vector spaces it must also be surjective.
This being said, if you take $w \in U^{\top}$ and an arbitrary element of $U$, write it in the form $u = T(u')$, then $T(w)$ is also in $U^{\top}$ because $(T(w), T(u')) = (w,u') = 0$. Since $u = T(u')$ was arbitrary, we conclude $T(w) \in U^{\top}$ and $U^{\top}$ is $T$-invariant.
Hope that helps,