Question

Let A be a \(3 \times 3\) real matrix with det(\(A + i I\)) = 0, where \(i = \sqrt{-1}\) and I is the \(3 \times 3\) identity matrix. If det(A) = 3, then the trace of \(A^2\) is ..................

Hint:

For any real matrix, complex eigenvalues always appear in conjugate pairs. This is a crucial property for solving many problems involving eigenvalues of real matrices. Also, remember the fundamental relationships: det(A) = product of eigenvalues, and trace(A) = sum of eigenvalues.

Answer 1

Step 1: Understanding the Concept:
This problem connects several key concepts in linear algebra: eigenvalues, determinant, trace, and properties of real matrices. The condition det(\(A - \lambda I\)) = 0 is the definition of \(\lambda\) being an eigenvalue of A.
Step 2: Key Formula or Approach:
1. Use the condition det(\(A + iI\)) = 0 to find one of the eigenvalues of A.
2. Use the property that for a real matrix A, if \(\lambda\) is a complex eigenvalue, then its complex conjugate \(\bar{\lambda}\) is also an eigenvalue.
3. Use the property that the determinant of a matrix is the product of its eigenvalues: det(A) = \(\lambda_1 \lambda_2 \lambda_3\).
4. Use the property that if \(\lambda\) is an eigenvalue of A, then \(\lambda^k\) is an eigenvalue of \(A^k\).
5. Use the property that the trace of a matrix is the sum of its eigenvalues: trace(\(A^2\)) = \(\lambda_1^2 + \lambda_2^2 + \lambda_3^2\).
Step 3: Detailed Explanation or Calculation:
Finding the eigenvalues of A:
The given condition is det(\(A + iI\)) = 0, which can be written as det(\(A - (-i)I\)) = 0. By the definition of eigenvalues, this means that \(\lambda_1 = -i\) is an eigenvalue of A.
Since A is a real matrix, its characteristic polynomial has real coefficients. Therefore, complex roots must come in conjugate pairs. If \(-i\) is an eigenvalue, then its complex conjugate, \(\overline{-i} = i\), must also be an eigenvalue. So, \(\lambda_2 = i\).
Let the third eigenvalue be \(\lambda_3\). The determinant of A is the product of its eigenvalues.
\[ \det(A) = \lambda_1 \lambda_2 \lambda_3 \] We are given det(A) = 3. \[ 3 = (-i)(i)(\lambda_3) = (-i^2)(\lambda_3) = (1)(\lambda_3) \] Therefore, \(\lambda_3 = 3\).
The eigenvalues of A are \(-i, i, 3\).
Finding the trace of \(A^2\):
If the eigenvalues of A are \(\lambda_1, \lambda_2, \lambda_3\), then the eigenvalues of \(A^2\) are \(\lambda_1^2, \lambda_2^2, \lambda_3^2\).
The eigenvalues of \(A^2\) are: \[ \lambda_1^2 = (-i)^2 = i^2 = -1 \] \[ \lambda_2^2 = (i)^2 = -1 \] \[ \lambda_3^2 = (3)^2 = 9 \] The trace of \(A^2\) is the sum of its eigenvalues. \[ \text{trace}(A^2) = \lambda_1^2 + \lambda_2^2 + \lambda_3^2 = (-1) + (-1) + 9 = 7 \] Step 4: Final Answer:
The trace of \(A^2\) is 7.
Step 5: Why This is Correct:
The solution correctly deduces the three eigenvalues of the matrix A by using the given information and fundamental properties of real matrices. Then, it correctly finds the eigenvalues of \(A^2\) and calculates their sum to find the trace, arriving at the correct answer of 7.