Suppose $A \in \mathbb{R}^{n \times n}$. $A$ is dissipative if $A+A^T \prec 0$. How do I show that $A$ is similar to a dissipative matrix?
I started by showing that the eigenvalue of $A$ is real and less than zero but I can't think of any way to proceed ahead. I think it would be helpful to use some type of matrix decomposition, perhaps Schur decomposition? Please advise.
I can provide a dynamic systems view on this using continuous dynamic systems in state-space representation. Particularly, a system represented by the equation $$ \dfrac{\mathbb{d}\vec{x}(t)}{\mathbb{d}t} = \mathbf{A} \vec{x}(t) ,\; \mathbf{A} \in \mathbb{R}^{n \times n} ,\; \vec{x} \in \mathbb{R}^n . $$ It can be shown that the time response of such system is given by $$ \vec{x}(t) = e^{\mathbf{A}t} \vec{x}(0) ,\; t \geq 0 , $$ where $\vec{x}(0)$ are the initial conditions at time $t=0$.
Let $$ \mathbf{V} = \begin{bmatrix} \vec{v_1} & \vec{v_2} & \cdots & \vec{v_n} \end{bmatrix} , $$ where $\{\vec{v_1}, \vec{v_2}, \dotsc, \vec{v_n}\}$ are the eigenvectors of $\mathbf{A}$. Then, using $\mathbf{V}$ as a transformation matrix $$ \mathbf{\Lambda} = \mathbf{V}^{-1} \mathbf{A} \mathbf{V} . $$ $\mathbf{\Lambda}$ is said to be the Jordan form of $\mathbf{A}$. It looks like $$ \mathbf{\Lambda} = \begin{bmatrix} \lambda_1 & 0 & 0 & 0 & 0 & \cdots & 0 \\ 0 & \lambda_2 & 1 & 0 & 0 & \cdots & 0 \\ 0 & 0 & \lambda_2 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 0 & \lambda_2 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 0 & \lambda_5 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \\ 0 & 0 & 0 & 0 & 0 & \cdots & \lambda_n \\ \end{bmatrix} , $$ where $\{\lambda_1, \lambda_2, \dotsc, \lambda_n\}$ are the eigenvalues of $\mathbf{A}$ and $\lambda_2 = \lambda_3 = \lambda_4$ are repeated eigenvalues. Notice the $1$s above the repeated $\lambda_2$; they may or may not appear above repeated eigenvalues. Then $$ e^{\mathbf{\Lambda}t} = \begin{bmatrix} e^{\lambda_1 t} & 0 & 0 & 0 & 0 & \cdots & 0 \\ 0 & e^{\lambda_2 t} & \dfrac{t e^{\lambda_2 t}}{1!} & \dfrac{t^2 e^{\lambda_2 t}}{2!} & 0 & \cdots & 0 \\ 0 & 0 & e^{\lambda_2 t} & \dfrac{t e^{\lambda_2 t}}{1!} & 0 & \cdots & 0 \\ 0 & 0 & 0 & e^{\lambda_2 t} & 0 & \cdots & 0 \\ 0 & 0 & 0 & 0 & e^{\lambda_5 t} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \\ 0 & 0 & 0 & 0 & 0 & \cdots & e^{\lambda_n t} \\ \end{bmatrix} $$
Also, it can be shown that $$ e^{\mathbf{A}t} = \mathbf{V} e^{\mathbf{\Lambda}t} \mathbf{V}^{-1} $$ which applied to the system time response, gives $$ \vec{x}(t) = \mathbf{V} e^{\mathbf{\Lambda}t} \mathbf{V}^{-1} \vec{x}(0) ,\; t \geq 0 $$
Suppose $\vec{x}(0) \neq \vec{0}$. Then, $\vec{x}(t) \to \vec{0} \iff e^{\mathbf{\Lambda}t} \to \mathbf{0}$; that is, all states decay (or dissipate energy) if and only if every element in $e^{\mathbf{\Lambda}t}$ goes to zero over time. All elements are either already zero or $t^k e^{\lambda t},\; k \geq 0$, which tends to zero if and only if $\Re(\lambda) < 0$; that is, the real part of $\lambda$ is negative.
Thus, $\mathbf{A}$ is dissipative if and only if the real part of all of its eigenvalues are negative, which implies it is negative definite, which implies $\mathbf{A}+\mathbf{A}^T < \mathbf{0}$.