Stability of the Zero Solution - summary

Question

I was studying the stability of the zero solutions to an IVP and I came across a list of bullets with True/False questions. I tried to answer and give an argument for my answer but I am not sure if my reasoning is good as well as the correctness of my answers, I would appreciate any help or comment:

So if we consider IVP:

$$ \dot{x}=f(x); $$

$$ x(t_0)=x_0 $$

with $f:\mathbb{R^n}\rightarrow\mathbb{R^n}$, $ \; t_0 \in \mathbb{R}$, $ \; x_0\in\mathbb{R^n}$.

1. If $f = Ax$ with $A$ real $n\times n$ matrix, the zero solution is stable if $\lambda \leq0$ for all eigenvalues $\lambda$ of $A$. My Answer: TRUE, because $\lambda \leq 0$.(Are geometric and algebraic multiplicity important here?)
2. If $f = Ax$ with $A$ real $n\times n$ matrix in diagonal form, the zero solution is stable if $\lambda \leq0$ for all eigenvalues $\lambda$ of $A$. My Answer: TRUE, because $\lambda \leq 0$.Same question as for Nr. 1?
3. If $f = Ax$ with $A$ real $n\times n$ matrix in diagonal form, the zero solution is asymptotically stable if $\lambda \leq0$ for all eigenvalues $\lambda$ of $A$. My Answer: FALSE, because $\lambda \leq 0$, $\lambda$ should be $<0.$
4. If $f = Ax$ with $A$ real $n\times n$ matrix, the zero solution is asymptotically stable if $\lambda < 0$ for all eigenvalues $\lambda$ of $A$. My Answer: TRUE, because $\lambda < 0$.
5. If $f = Ax$ with $A$ real $n\times n$ matrix, the zero solution is unstable if $A$ posses nonreal eigenvalues My Answer: FALSE, because we check if the real part of $\lambda$ is $\leq 0$.
6. If $f = Ax$ with $A$ real $n\times n$ symmetrical matrix, the zero solution is stable if $\lambda \leq 0$ for all eigenvalues $\lambda$ of $A$. My Answer: TRUE, because I don´t see how symmetry would affect stability.
7. If $f = Ax$ with $A$ real $n\times n$ symmetrical matrix, the zero solution is asymptotically stable if $\lambda < 0$ for all eigenvalues $\lambda$ of $A$. My Answer: TRUE, because I don´t see how symmetry would affect stability again.