e know that :
• tr(A) = P
i λi
• det(A) = Q
i λi
Prove it for the special case of a symmetric matrix A

To prove the given statements for a symmetric matrix \( A \), we need to show two things:

1. \( \text{tr}(A) = \sum_{i} \lambda_i \), where \( \lambda_i \) are the eigenvalues of \( A \).
2. \( \det(A) = \prod_{i} \lambda_i \), where \( \lambda_i \) are the eigenvalues of \( A \).

These properties hold for any matrix that is diagonalizable, but we will focus specifically on the symmetric case, where these properties are simpler to demonstrate because symmetric matrices have real eigenvalues and can always be diagonalized using an orthogonal matrix. Let's prove these step by step.

### Key properties of symmetric matrices:
- **Diagonalizability**: A symmetric matrix \( A \) can be diagonalized as \( A = U \Lambda U^T \), where \( U \) is an orthogonal matrix (i.e., \( U^T U = I \)) and \( \Lambda \) is a diagonal matrix whose diagonal entries are the eigenvalues \( \lambda_1, \lambda_2, \dots, \lambda_n \) of \( A \).
  
Thus, \( A = U \Lambda U^T \) implies \( A \) is similar to the diagonal matrix \( \Lambda \), meaning \( A \) and \( \Lambda \) share the same eigenvalues.

### 1. **Trace of a symmetric matrix \( A \):**

The trace of a matrix is defined as the sum of its diagonal elements. Since \( A \) is symmetric, it is diagonalizable as \( A = U \Lambda U^T \), and the trace of a matrix is invariant under similarity transformations. Therefore:
\[
\text{tr}(A) = \text{tr}(U \Lambda U^T) = \text{tr}(\Lambda)
\]
The trace of the diagonal matrix \( \Lambda \) is simply the sum of its diagonal elements, which are the eigenvalues of \( A \). Thus:
\[
\text{tr}(A) = \lambda_1 + \lambda_2 + \dots + \lambda_n = \sum_{i} \lambda_i
\]
This proves that the trace of a symmetric matrix \( A \) is equal to the sum of its eigenvalues.

### 2. **Determinant of a symmetric matrix \( A \):**

The determinant of a matrix is also invariant under similarity transformations. Using the diagonalization \( A = U \Lambda U^T \), the determinant of \( A \) is:
\[
\det(A) = \det(U \Lambda U^T) = \det(U) \det(\Lambda) \det(U^T)
\]
Since \( U \) is an orthogonal matrix, \( \det(U) = \pm 1 \), and \( \det(U^T) = \det(U)^{-1} = \pm 1 \). Therefore, \( \det(A) = \det(\Lambda) \).

The determinant of the diagonal matrix \( \Lambda \) is the product of its diagonal entries, which are the eigenvalues \( \lambda_1, \lambda_2, \dots, \lambda_n \). Hence:
\[
\det(A) = \lambda_1 \lambda_2 \dots \lambda_n = \prod_{i} \lambda_i
\]
This proves that the determinant of a symmetric matrix \( A \) is equal to the product of its eigenvalues.

### Conclusion:
For a symmetric matrix \( A \), we have shown that:
- \( \text{tr}(A) = \sum_{i} \lambda_i \),
- \( \det(A) = \prod_{i} \lambda_i \),
where \( \lambda_i \) are the eigenvalues of \( A \).

Let me know if you'd like further details or explanations!

Here are 5 questions to extend your understanding:
1. How does the trace and determinant property change for non-symmetric matrices?
2. What role does the orthogonal matrix \( U \) play in the diagonalization of symmetric matrices?
3. Can you extend these results to complex symmetric matrices, and how do the properties of eigenvalues differ?
4. Why is diagonalizability so crucial for proving these properties for symmetric matrices?
5. How do the properties of positive definite symmetric matrices relate to their eigenvalues?

**Tip**: Remember that the trace of any square matrix is invariant under similarity transformations, which is why it's so useful in this context!

We know that: 
• tr(A) = ∑ λi 
• det(A) = ∏ λi 
Prove it for the special case of a symmetric matrix A.

Explore a step-by-step proof for the trace and determinant properties of symmetric matrices, where the trace equals the sum of eigenvalues and the determinant equals the product of eigenvalues. This explanation involves key concepts from linear algebra, including matrix diagonalization, eigenvalues, and orthogonal matrices.

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