Question

If $A^2 = 0$, then:

• A. $e^{At} = I$
• B. $e^{At} = I + At$
• C. $e^{At} = At$
• D. Infinite series required

Hint:

For any nilpotent matrix of order \(n\) (where \(A^n = 0\)), the matrix exponential \(e^{At}\) is a polynomial in \(t\) of degree at most \(n-1\).
This property is widely used in solving linear differential equations with nilpotent system matrices.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the state transition matrix \(e^{At}\) (matrix exponential) for a given matrix \(A\) that is nilpotent of order 2, meaning \(A^2 = 0\).

Step 2: Key Formula or Approach:
The matrix exponential \(e^{At}\) is defined by the Taylor series expansion:
\[ e^{At} = \sum_{k=0}^{\infty} \frac{(At)^k}{k!} = I + At + \frac{A^2 t^2}{2!} + \frac{A^3 t^3}{3!} + \dots \] where \(I\) is the identity matrix.

Step 3: Detailed Explanation:

• We are given the condition \(A^2 = 0\) (where \(0\) is the zero matrix).

• Since \(A^2 = 0\), all higher powers of \(A\) must also be zero:
\[ A^3 = A^2 \cdot A = 0 \cdot A = 0 \] \[ A^4 = A^3 \cdot A = 0 \cdot A = 0 \]
• In general, \(A^k = 0\) for all \(k \ge 2\).

• Substituting these values into the Taylor series expansion of \(e^{At}\):
\[ e^{At} = I + At + \frac{0 \cdot t^2}{2!} + \frac{0 \cdot t^3}{3!} + \dots \] \[ e^{At} = I + At \]
• Thus, the infinite series terminates after the second term, and the exact representation is \(e^{At} = I + At\).

Step 4: Final Answer:
The matrix exponential is \(e^{At} = I + At\).