Finding all symmetrical matrices, such that zero solution is stable

Question

Determine all symmetrical real $n\times n $ matrices $A$ with the property, such that the zero solution of ODE $\dot{x}=A^2x$ is stable.

I am confused about how to approach this problem and I would appreciate any help.

Answer 1

The only such matrix is $A=0$. Any symmetric real matrix $A$ can be diagonalized by an orthogonal matrix $U$: $$ A=U^T D U,\quad D=\left(\begin{array}{ccc} \lambda_1&\ldots&0\\ \vdots&\ddots&\vdots\\ 0&\ldots&\lambda_n\\ \end{array}\right),\quad U U^T=I, $$ where $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $A$, $\lambda_1,\ldots,\lambda_n\in\mathbb R$. Then $$ A^2=U^T D UU^T D U=U^T D^2 U =U^T\left(\begin{array}{ccc} \lambda_1^2&\ldots&0\\ \vdots&\ddots&\vdots\\ 0&\ldots&\lambda_n^2\\ \end{array}\right)U $$ This means that the eigenvalues of $A^2$ are $\lambda_1^2,\ldots,\lambda_n^2$. If at least one of the eigenvalues of $A$, $\lambda_i$, is not equal to zero, then the corresponding eigenvalue of $A^2$ is $\lambda_i^2>0$ and the system $\dot x=A^2x$ is unstable. Thus, $\lambda_1=\ldots=\lambda_n=0$ and $A=0$ there is only one way for the system $\dot x=A^2x$ to be stable.