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The problem involves a matrix \( A \) that satisfies the property \( A^T = -A \), where \( A^T \) denotes the transpose of \( A \). The questions are as follows:

### a) If \( A \) has an eigenvalue \( \lambda \), is \(-\lambda\) also an eigenvalue of \( A \)? Provide reasoning.

**Answer:**

Yes, if \( \lambda \) is an eigenvalue of \( A \), then \(-\lambda\) is also an eigenvalue of \( A \).

**Reasoning:**

If \( \lambda \) is an eigenvalue of \( A \), there exists a non-zero vector \( v \) such that \( A v = \lambda v \). Taking the transpose of both sides, we get:
\[
v^T A^T = \lambda v^T
\]
Given \( A^T = -A \), this equation becomes:
\[
v^T (-A) = \lambda v^T
\]
which simplifies to:
\[
v^T A = -\lambda v^T
\]
This implies that \( v^T \) is a left eigenvector of \( A \) corresponding to the eigenvalue \(-\lambda\). Since the eigenvalues of a matrix and its transpose are the same, \(-\lambda\) is also an eigenvalue of \( A \).

### b) Prove that for any real vector \( \alpha \), \( \alpha^T A \alpha = 0 \).

**Answer:**

We know that \( A^T = -A \). Let \( \alpha \) be any real vector. Consider the expression \( \alpha^T A \alpha \). Taking the transpose of this expression:
\[
(\alpha^T A \alpha)^T = \alpha^T A^T \alpha
\]
Since \( A^T = -A \), we have:
\[
\alpha^T A^T \alpha = \alpha^T (-A) \alpha = -\alpha^T A \alpha
\]
Thus, \( \alpha^T A \alpha = -\alpha^T A \alpha \), which implies:
\[
2 \alpha^T A \alpha = 0
\]
Therefore, \( \alpha^T A \alpha = 0 \).

### c) Prove that if the eigenvalue of \( A \) is non-zero, then it is purely imaginary.

**Answer:**

Suppose \( \lambda \) is an eigenvalue of \( A \) and let \( v \) be the corresponding eigenvector such that \( A v = \lambda v \). Taking the transpose:
\[
v^T A^T = \lambda v^T
\]
Using \( A^T = -A \), this becomes:
\[
v^T (-A) = \lambda v^T
\]
which simplifies to:
\[
-\lambda v^T = \lambda v^T
\]
Since \( v \) is a non-zero vector, we must have \( \lambda = -\lambda \), which implies \( \lambda = 0 \) or \( \lambda \) is purely imaginary (i.e., \( \lambda = i\mu \) for some real \( \mu \)).

### d) Prove that the determinant of \( A \) is not negative.

**Answer:**

The determinant of an antisymmetric matrix of odd order is always zero, and for an even-order matrix, it is non-negative. Since \( A \) is antisymmetric (because \( A^T = -A \)), and its eigenvalues are either zero or purely imaginary, the determinant, which is the product of the eigenvalues, cannot be negative. 

Thus, \(\text{det}(A)\) is either zero or positive, and hence it is not negative.

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Would you like more details on any part of the explanation? Here are eight related questions to explore further:

1. What is the significance of an antisymmetric matrix?
2. How do eigenvalues relate to the determinant of a matrix?
3. Can a matrix have both real and imaginary eigenvalues?
4. What is the geometric interpretation of purely imaginary eigenvalues?
5. How does the transpose operation affect the properties of a matrix?
6. What are the implications of having a zero determinant?
7. How would the results change if the matrix was symmetric instead of antisymmetric?
8. Can you prove that the trace of an antisymmetric matrix is always zero?

**Tip:** When dealing with properties of matrices, understanding the implications of symmetric and antisymmetric matrices is crucial as they often reveal deeper insights into the matrix's structure and behavior.

Explore the properties of antisymmetric matrices and eigenvalues with detailed explanations and proofs. Learn about the relationship between eigenvalues and matrix transposes, and the determinant properties of antisymmetric matrices. Suitable for advanced undergraduate students and anyone interested in linear algebra.

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