Let's address the problems in detail:

A. Prove that there exists a basis of $V$ over $F$ consisting of characteristic vectors of $T$:

If $T \in A_F(V)$ has $n$-distinct characteristic roots in $F$, then $T$ is a linear operator on $V$ with eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_n$, which are distinct. This means the minimal polynomial of $T$ splits into distinct linear factors over $F$: $m_T(x) = (x - \lambda_1)(x - \lambda_2)\cdots(x - \lambda_n).$

Since $m_T(x)$ splits into distinct factors, $V$ can be decomposed into a direct sum of eigenspaces corresponding to each $\lambda_i$: $V = E_{\lambda_1} \oplus E_{\lambda_2} \oplus \cdots \oplus E_{\lambda_n},$ where $E_{\lambda_i} = \ker(T - \lambda_iI)$.

Each $E_{\lambda_i}$ is spanned by eigenvectors corresponding to $\lambda_i$. By selecting a basis for each $E_{\lambda_i}$, and combining them, we obtain a basis for $V$ that consists entirely of eigenvectors of $T$, as required.

B1. Prove that $T$ is invertible if and only if the constant term of the minimal polynomial is not zero:

Let the minimal polynomial of $T$ be: $m_T(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0,$ where $a_0$ is the constant term.

(1) $T$ is invertible $\implies a_0 \neq 0$:

If $T$ is invertible, $0$ is not an eigenvalue of $T$. Therefore, $x = 0$ is not a root of $m_T(x)$. This implies that the minimal polynomial $m_T(x)$ does not contain $x$ as a factor, so $a_0 \neq 0$.

(2) $a_0 \neq 0 \implies T$ is invertible:

If $a_0 \neq 0$, then $m_T(x)$ does not have $x = 0$ as a root. Since the minimal polynomial includes all eigenvalues of $T$, $0$ is not an eigenvalue. Thus, $T$ is invertible.

B2. Prove that $V = A \oplus B$ for some subspaces $A, B$ of $V$ if $T$ is idempotent:

If $T \in A_F(V)$ is idempotent, then $T^2 = T$. This implies that $T$ is a projection operator, and $V$ can be decomposed as: $V = \ker(T) \oplus \operatorname{im}(T),$ where:

• $\ker(T)$ is the null space of $T$, consisting of all vectors $v$ such that $T(v) = 0$,
• $\operatorname{im}(T)$ is the image of $T$, consisting of all vectors $v$ such that $v = T(v)$.

Thus, setting $A = \ker(T)$ and $B = \operatorname{im}(T)$, we have: $V = A \oplus B.$

This completes the proof.

Would you like a deeper explanation or examples for any part? Here are some related questions:

1. What is the relationship between the minimal polynomial and characteristic polynomial in proving basis existence?
2. How does the splitting of $m_T(x)$ into distinct linear factors ensure diagonalizability?
3. Could you provide examples of idempotent operators in specific vector spaces?
4. What happens when $T$ has repeated characteristic roots?
5. Can $T$ fail to be invertible even if all eigenvalues are nonzero?

Tip: To verify if a basis consists of eigenvectors, ensure that $T(v) = \lambda v$ holds for every basis vector $v$.