Problem 1706

### Part (a)
1. Given that $A \in \mathcal{M}_n(\mathbb{R}^*)$ and $A \cdot A^T = I_n$, we know that $A$ is an orthogonal matrix.
2. The eigenvalues of an orthogonal matrix $A$ lie on the unit circle in the complex plane, meaning if $\lambda_i$ is an eigenvalue of $A$, then $|\lambda_i| = 1$.
3. The trace of $A$, denoted $\text{Tr}(A)$, is the sum of its eigenvalues: $\text{Tr}(A) = \sum_{i=1}^n \lambda_i$.
4. Using the triangle inequality, we have:
$$|\text{Tr}(A)| = \left| \sum_{i=1}^n \lambda_i \right| \leq \sum_{i=1}^n |\lambda_i| = \sum_{i=1}^n 1 = n.$$
5. Therefore, we conclude that:
$$|\text{Tr}(A)| \leq n.$$

### Part (b)
1. Given that $n$ is odd, we need to show that $\det(A^2 - I_n) = 0$.
2. Since $A$ is orthogonal, its eigenvalues $\lambda_i$ satisfy $|\lambda_i| = 1$. Therefore, the eigenvalues of $A$ are either $1$, $-1$, or complex numbers of unit modulus.
3. For an orthogonal matrix $A$ with odd dimension $n$, there must be at least one real eigenvalue. This is because the complex eigenvalues of $A$ come in conjugate pairs, and since $n$ is odd, there must be an unpaired eigenvalue which must be real.
4. The real eigenvalues of $A$ can only be $1$ or $-1$ because $|\lambda_i| = 1$.
5. Therefore, $A$ must have at least one eigenvalue equal to $1$ or $-1$.
6. If $\lambda = 1$ is an eigenvalue of $A$, then $\lambda^2 - 1 = 0$. Similarly, if $\lambda = -1$ is an eigenvalue of $A$, then $\lambda^2 - 1 = 0$.
7. This implies that $A^2 - I_n$ has at least one zero eigenvalue, making $\det(A^2 - I_n) = 0$.

Proof 1:
1. Since $n$ is odd, $A$ must have at least one real eigenvalue, which is either $1$ or $-1$.
2. Therefore, at least one of the matrices $A - I$ or $A + I$ is singular.
3. We observe that:
$$\det(A^2 - I) = \det((A - I)(A + I)).$$
4. Since at least one of $A - I$ or $A + I$ is singular, their product is also singular, implying:
$$\det(A^2 - I) = 0.$$

Proof 2:
1. We start with:
$$\det(A^2 - I) = \det((A - I)(A + I)).$$
2. Using the property of determinants and the fact that $A$ is orthogonal ($A^T = A^{-1}$), we have:
$$\det(A^2 - I) = \det(A - I) \cdot \det(A + I).$$
3. Since $A$ is orthogonal, $A^T = A^{-1}$, and thus:
$$\det(A + I) = \det(A^T + I) = \det(I + A - A^T - I) = \det(A - A^T).$$
4. On one hand, $\det(A - A^T) = \det(A^T - A)$.
5. On the other hand, since $\det(A - A^T) = (-1)^n \det(A^T - A)$ and $n$ is odd, we have:
$$\det(A - A^T) = -\det(A^T - A).$$
6. This implies:
$$\det(A - A^T) = 0.$$
7. Therefore, we conclude that:
$$\det(A^2 - I) = 0.$$

The final answer is $\boxed{ |\text{Tr}(A)| \le n }$ and $\det(A^2 - I_n) = 0$ if $n$ is odd.