91Ó°ÊÓ

A matrix A is diagonalizable if there exists an invertible matrix P and a diagonal matrix D such that \(A = PDP^{-1}\), where D is a diagonal matrix formed from the eigenvalues of the matrix A on its diagonal. In other words, A has sufficient linearly independent eigenvectors to be diagonalizable.

The matrix exponential \(e^A\) is defined as the infinite sum of the power series: \[e^A = I + A + \frac{1}{2!}A^2 + \frac{1}{3!}A^3 + \cdots.\] Where \(I\) is the identity matrix. The inverse of a matrix exponential, denoted as \(e^{-A}\), can be found by simply replacing \(A\) with \(-A\) in the infinite sum definition.

Since A is a diagonalizable matrix, we have \(A = PDP^{-1}\). Let's find the exponential of A now. The exponential of A is given by \[e^A = Pe^DP^{-1}.\] The exponential of the diagonal matrix D is simply the matrix with the exponential of each of its diagonal elements, i.e., if \(D = diag(\lambda_1, \lambda_2, \ldots, \lambda_n)\), then \(e^D = diag(e^{\lambda_1}, e^{\lambda_2}, \ldots, e^{\lambda_n})\).

We need to show that the matrix exponential \(e^A\) is nonsingular. A matrix is nonsingular if it has an inverse. Let's find the inverse of \(e^A\): \[e^{-A} = Pe^{-D}P^{-1},\] Where \(e^{-D} = diag(e^{-\lambda_1}, e^{-\lambda_2}, \ldots, e^{-\lambda_n})\). Now, we show that \(e^A\) multiplied by its inverse gives the identity matrix: \begin{align*} (e^A)(e^{-A}) &= (Pe^DP^{-1})(Pe^{-D}P^{-1}) \\ &= Pe^D(e^{-D}P^{-1}P)P^{-1} \\ &= Pe^De^{-D}P^{-1}. \end{align*} Since \(e^D\) and \(e^{-D}\) are diagonal matrices with positive elements (due to the exponential function), their product results in a diagonal matrix with positive elements as well. Thus, the product of \(e^D\) and \(e^{-D}\) is the identity matrix: \[e^De^{-D} = I.\] Plugging this back in, we get: \[(e^A)(e^{-A}) = PIP^{-1} = PP^{-1} = I.\] This shows that \(e^A\) is nonsingular, since it has an inverse. This concludes the proof that the matrix exponential \(e^A\) is nonsingular for any diagonalizable matrix A.