Question

If \(A\) is a square matrix such that \(A^2 = A\) and \((I + A)^n = I + \lambda A\), then \(\lambda\) is equal to

• A. \(2n - 1\)
• B. \(2^n - 1\)
• C. \(2n + 1\)
• D. None of these

Hint:

For an idempotent matrix \(A\), \(A^k = A\) for all positive integers \(k\).

Answer 1

The Correct Option is B

To solve the given problem, we need to analyze the properties of the matrix \( A \) and the given equations.

1. First, note that \( A \) is a square matrix such that \( A^2 = A \). This implies that \( A \) is an idempotent matrix. An important property of idempotent matrices is that their eigenvalues are either \( 0 \) or \( 1 \).
2. The equation \((I + A)^n = I + \lambda A\) holds for some scalar \(\lambda\). We need to find the value of \(\lambda\).
3. Let's consider the eigenvalues of \( I + A \). If the eigenvalue of \( A \) is \( 0 \), then the corresponding eigenvalue of \( I + A \) is \( 1 \). If the eigenvalue of \( A \) is \( 1 \), then the corresponding eigenvalue of \( I + A \) is \( 2 \).
4. For the matrix \( (I + A)^n \), each eigenvalue raised to the power of \( n \) also forms an eigenvalue of the resulting matrix. Therefore, the eigenvalues of \((I + A)^n\) will be \(1^n = 1\) and \(2^n\).
5. Since \((I + A)^n = I + \lambda A\), compare eigenvalues:
   • For eigenvalue corresponding to \(I\), eigenvalue is \(1\).
   • For eigenvalue corresponding to \(A\), eigenvalue is \(\lambda \cdot 1 = 2^n\).

Therefore, the value of \(\lambda\) for which the equation \((I + A)^n = I + \lambda A\) holds true is \(\boxed{2^n - 1}\).