Question

Let \( A \in M_2({C}) \) be given by \[ A = \begin{bmatrix} 0 & 2
2 & 0 \end{bmatrix}. \] Let \( T: M_2({C}) \to M_2({C}) \) be the linear transformation given by \( T(B) = AB \). The characteristic polynomial of \( T \) is:

• A. \( \lambda^4 - 8\lambda^2 + 16 \)
• B. \( \lambda^4 - 4 \)
• C. \( \lambda^4 - 2 \)
• D. \( \lambda^4 - 16 \)

Hint:

For matrix transformations, use eigenvalues to determine the characteristic polynomial.

Answer 1

The Correct Option is A

Step 1: Understanding the transformation. The matrix \( A \) acts as a linear map \( T \) on \( M_2({C}) \), where the characteristic polynomial of \( T \) is determined by the eigenvalues of \( A \). Step 2: Calculating the eigenvalues. The eigenvalues of \( A \) are \( \pm 2 \), so the characteristic polynomial of \( A \) is: \[ \lambda^2 - 4. \] Since \( T \) acts on \( M_2({C}) \), the eigenvalues of \( T \) are \( \pm 2, \pm 2 \), and the characteristic polynomial becomes: \[ (\lambda^2 - 4)^2 = \lambda^4 - 8\lambda^2 + 16. \] Step 3: Conclusion. The characteristic polynomial is \( {(1)} \lambda^4 - 8\lambda^2 + 16 \).