Question: If A = $\begin{bmatrix} \alpha & \alpha \\ \alpha & \alpha \end{bmatrix}$. Prove that $e^A = e^{\alpha} \begin{bmatrix} cosh\alpha&sinh\alpha \\ sinh\alpha&cosh\alpha \end{bmatrix}$
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Subject: Applied Mathematics 4

Topic: Matrix Theory

Difficulty: Medium

m4e(34) • 383 views
 modified 12 months ago  • written 16 months ago by
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For characteristic equation |A - $\lambda$ I| = 0

$$\begin{vmatrix} \alpha - \lambda & \alpha \\ \alpha &\alpha - \lambda \end{vmatrix} = 0$$

$(\alpha - \lambda) ^2 - \alpha^2 = 0 \\ \lambda^2 - 2\alpha\lambda + \alpha^2 - \alpha^2 = 0 \\ \lambda(\lambda-2\alpha) = 0 \\ \lambda = 0; \lambda = 2 \alpha$

Let,

$\phi(A) = e^{A} \\ \phi(A) = aA + bI \\ \phi(\lambda) = e^{\lambda} \\ \phi(\lambda) = a\lambda + b \\ \phi(0) = e^0 = 1 \\ \phi(2\alpha) = e^{2\alpha)}$

case(i) $\lambda$ = 0

$\phi(0) = a(0) + b \\ 1 = 0 + b \\ b = 1$

case(ii) $\lambda = 2 \alpha$

$\phi(2\alpha) = a(2\alpha) + b \\ e^{2\alpha} = a(2\alpha) + 1 \\ a = \frac{e^{2\alpha} - 1}{2\alpha}$

$\phi(A) = aA + bI \\ = \frac{e^{2\alpha} - 1}{2\alpha} \begin{bmatrix} \alpha & \alpha \\ \alpha & \alpha \end{bmatrix} + 1 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ = \begin{bmatrix} \frac{e^{2\alpha} - 1}{2} & \frac{e^{2\alpha} - 1}{2} \\ \frac{e^{2\alpha} - 1}{2} & \frac{e^{2\alpha} - 1}{2} \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ = \begin{bmatrix} \frac{e^{2\alpha} + 1}{2} & \frac{e^{2\alpha} - 1}{2} \\ \frac{e^{2\alpha} - 1}{2} & \frac{e^{2\alpha} + 1}{2} \end{bmatrix} \\ = \begin{bmatrix} e^\alpha cosh\alpha & e^\alpha sinh\alpha \\ e^\alpha sinh\alpha & e^\alpha cosh\alpha \end{bmatrix} \\ = e^\alpha \begin{bmatrix} cosh\alpha & sinh\alpha \\ sinh\alpha & cosh\alpha \end{bmatrix}$