Given $\dot y = Ly$ with $y^{(0)} \neq \hat 0$, determine the behavior of the linear system of ODE's. We can express $y(0)$ as the linear combination, $$y(0) = \sum_{i = 1}^{n}\alpha_iv_i$$ where $\alpha_i$ is a scalar. Then we have a closed form for $y(t)$, $$y(t) = \sum_{i = 1}^{n} \alpha_i v_i e^{\lambda_it}$$ Consider the following cases for $L$ with First, let $L = -(A^TA)$. The matrix $A^TA$ is symmetric positive definite so it has all strictly positive real eigenvalues. Hence, $L$ has strictly negative real eigenvalues. Second, let $L = -(A^T + A)$. As a result of $A^T$ and $A$ having the same characteristic polynomial. $A^T, A$ have the same eigenvalues. So the eigenvalues of $L$ are strictly negative but double in magnitude of $(A^T A). Third, let$L = -(A^T - A)$. As stated above, eigenvalues are equal in each matrix so the eigenvalues of$L$must be$0\$.