Question

Consider a diagonalizable$\mathbf{3}\mathbf{\times }\mathbf{3}$matrix A such that the zero state is a stable equilibrium solution of the system$\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{A}\overrightarrow{\mathbf{x}}$. What can you sayabout the determinant and the trace of A.

Short answer

The solution is$trA<0$ and$dtA>0$ .

Step-by-step solution

Explanation of the solution 

Consider a diagonalizable $3\times 3$matrix $A$such that the zero state is stable equilibrium solution of the system as follows.

$\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$

It is given that the system is $\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$stable if and only if the trace is less than 0 and the determinant is greater than zero.

Since, the system is stable.

Thus, $trA<0$and $dtA>0$.