The effect of negative eigenvalues for a matrix $A$ in a linear system Let $A$ be a $n$x $n$ matrix and  $x'=Ax$ be a linear system.

Prove if true, give a counter example if it is not.

(i) Suppose that all the eigenvalues of $A$ are distinct and have a negative real part. Then every solution of $x'=Ax$ satisfies $$|x(t)|\leq |x(s)|$$ for all $t&gt;s$.

(ii) Suppose that all the eigenvalues of $A$ have a negative real part. Then every solution of $x'=Ax$ satisfies $$|x(t)|\leq |x(s)|$$ for all $t&gt;s$.

(iii) Suppose $A$ is symmetric and all of its eigenvalue have a negative real part. Then every solution of $x'=Ax$ satisfies $$|x(t)|\leq |x(s)|$$ for all $t&gt;s$.

I would say that I should give a proof for (iii) since the eigenvalue a symmetric matrix is real and hence negative real which implies that $$|x(t)|=\sqrt{(c_1e^{\lambda_1 t}v_{1_{\lambda_1}}+...+c_ne^{\lambda_n t}v_{1_{\lambda_n}})^2+...+(c_1e^{\lambda_1 t}v_{n_{\lambda_1}}+...+c_ne^{\lambda_n t}v_{n_{\lambda_n}})^2} \leq \sqrt{(c_1e^{\lambda_1 s}v_{1_{\lambda_1}}+...+c_ne^{\lambda_n s}v_{1_{\lambda_n}})^2+...+(c_1e^{\lambda_1 s}v_{n_{\lambda_1}}+...+c_ne^{\lambda_n s}v_{n_{\lambda_n}})^2}= |x(s)|$$

for any $t&gt;s$, where $c=(c_1,...,c_n)$ referes to the constant for the general solutions and $v_{{\lambda_i}}=(v_{1_{\lambda_i}},...,v_{n_{\lambda_i}})$ refers to the eigenvectors of $\lambda_i$. 

So it holds for (iii), I can not see that is true for i or ii but I can not find an example for that, I would appreciate any help with that. Thank you in advance.
Assume that $A$ is real. For (i), there exists a real invertible matrix $P$ such that $P^{-1}AP$ is block-diagonal,i.e. has diagonal elements and/or diagonal blocks of the form $$\begin{bmatrix}a&amp;-b\\b&amp;a\end{bmatrix}$$ corresponding to complex eigenvalues $a \pm bi$ where $a$ and $b$ are real. Then if we define new variables $y_1,...,y_n$ by $$[x_1,...,x_n]^T=P[y_1,...,y_n]^T$$each $y_i=e^{at}(c \cos(bt)+ d \sin(bt))$. Thus, because each $a$ is negative, each $y_i$ is bounded and therefore each $x_i$ is also bounded. For (ii), here is a brief sketch. We put the matrix into block Jordan form where there are some 2x2 identity matrices along the super-diagonal. The difference from (i) is that when we solve for the $y_i$ we obtain expressions of the form $$ e^{at}f(t)(c \cos(bt)+ d \sin(bt)) $$ where $f$ is a polynomial, which you can show are also bbounded.
Assume that $A$ is real. For (i), there exists a real invertible matrix $P$ such that $P^{-1}AP$ is block-diagonal,i.e. has diagonal elements and/or diagonal blocks of the form $$\begin{bmatrix}a&amp;-b\\b&amp;a\end{bmatrix}$$ corresponding to complex eigenvalues $a \pm bi$ where $a$ and $b$ are real. Then if we define new variables $y_1,...,y_n$ by $$[x_1,...,x_n]^T=P[y_1,...,y_n]^T$$each $y_i=e^{at}(c \cos(bt)+ d \sin(bt))$. Thus, because each $a$ is negative, each $y_i$ is bounded and therefore each $x_i$ is also bounded. For (ii), here is a brief sketch. We put the matrix into block Jordan form where there are some 2x2 identity matrices along the super-diagonal. The difference from (i) is that when we solve for the $y_i$ we obtain expressions of the form $$ e^{at}f(t)(c \cos(bt)+ d \sin(bt)) $$ where $f$ is a polynomial, which you can show are also bbounded.