Let's tackle the questions one by one:

A. If $T \in A_F(V)$ has $n$-distinct characteristic roots in $F$, prove that there exists a basis of $V$ over $F$ that consists of characteristic vectors of $T$.

Proof:

1. Since $T$ has $n$-distinct characteristic roots, the minimal polynomial of $T$ over $F$ splits into $n$ distinct linear factors: $p_T(x) = (x - \lambda_1)(x - \lambda_2) \cdots (x - \lambda_n),$ where $\lambda_i$ are the eigenvalues of $T$.

2. For each $\lambda_i$, the eigenspace $E_{\lambda_i} = \ker(T - \lambda_i I)$ is nontrivial because $\lambda_i$ is a root of the characteristic polynomial.

3. Since the eigenvalues are distinct, the eigenspaces corresponding to different eigenvalues are linearly independent. Therefore: $V = \bigoplus_{i=1}^n E_{\lambda_i},$ which means $V$ is the direct sum of these eigenspaces.

4. In each eigenspace $E_{\lambda_i}$, choose a basis of characteristic (eigen)vectors for $T$.

5. Combining these bases for all eigenspaces forms a basis of $V$ over $F$ that consists entirely of characteristic vectors of $T$.

Hence, the proof is complete.

B. Do Any Two of the Following:

1. Let $T \in A_F(V)$, show that $T$ is invertible if and only if the constant term of the minimal polynomial for $T$ over $F$ is not 0.

Proof:

1. The minimal polynomial $p_T(x)$ of $T$ is the unique monic polynomial of least degree such that: $p_T(T) = 0.$

2. Let $p_T(x) = c_0 + c_1x + \cdots + x^d$, where $c_0$ is the constant term.

3. $T$ is invertible if and only if $0$ is not an eigenvalue of $T$, which implies that $\det(T) \neq 0$.

4. Note that $0$ is an eigenvalue of $T$ if and only if $x = 0$ is a root of $p_T(x)$. This happens if and only if $c_0 = 0$ (the constant term).

5. Thus, $T$ is invertible if and only if $c_0 \neq 0$.

2. If $T \in A_F(V)$ is idempotent, prove that $V = A \oplus B$ for some subspaces $A, B$ of $V$.

Proof:

1. Since $T$ is idempotent, we have: $T^2 = T.$

2. This implies that $T$ acts as a projection operator. Let $A = \text{Im}(T) = \{v \in V \mid T(v) = v\}$ and $B = \ker(T) = \{v \in V \mid T(v) = 0\}$.

3. We claim that $V = A \oplus B$. To verify:

   • Direct sum condition: If $v \in A \cap B$, then $T(v) = v$ (since $v \in A$) and $T(v) = 0$ (since $v \in B$). Thus, $v = 0$, so $A \cap B = \{0\}$.
   • Span condition: For any $v \in V$, decompose $v$ as: $v = T(v) + (v - T(v)),$ where $T(v) \in A$ and $v - T(v) \in B$. Hence, $V = A + B$.

4. Therefore, $V = A \oplus B$.

Let me know if you'd like detailed clarifications or further examples!

Follow-Up Questions for Practice

1. What is the relationship between the characteristic polynomial and the minimal polynomial of $T$?
2. Can you construct an example where $T$ has 3 distinct eigenvalues and determine its basis of eigenvectors?
3. How does the concept of eigenspaces help simplify transformations in a vector space?
4. Prove that idempotent operators are diagonalizable.
5. Explore the case where $T$ is not idempotent and how $V$ can still be decomposed.

Tip: Always start proofs with definitions and properties of the given operator for clarity and structure.