91Ó°ÊÓ

In order to show that \(e^A\) is symmetric and positive definite, we first calculate the matrix exponential for A, given by \(e^A = \sum_{k=0}^{\infty} \frac{A^k}{k!}\), where A is a symmetric \(n\times n\) matrix. We then find that \(e^A\) is symmetric by demonstrating that its transpose is equal to the original matrix exponential, i.e., \((e^A)^T = \sum_{k=0}^{\infty} \frac{(A^k)^T}{k!} = \sum_{k=0}^{\infty} \frac{A^k}{k!} = e^A\). To show that \(e^A\) is positive definite, we factor A using spectral decomposition as \(A = PDP^T\) and calculate the target expression for positive definiteness, \(v^T e^A v\), as \(v^T e^A v = \sum_{k=0}^{\infty} \frac{w^T D^k w}{k!}\), where \(w=Pv\). Lastly, we demonstrate that this expression is greater than zero for any non-zero vector v, showing that \(e^A\) is positive definite.