I have shown that for the given ODE system, that when we apply the forward Euler method to something like y′=Ay y(t_0)=y_0 t in (t_0,T] where A is diagonalizable; A=PDP^(−1), that the solution given by forward Euler is u_n=P(I+hD)^nP^(−1)u_0 where we consider the Forward Euler method to be y_(k+1)=y_k+hy′(x_k,y_k).

I have shown that for the given ODE system, that when we apply the forward Euler method to something like
$\begin{array}{rl}{\mathbf{y}}^{\prime }& =A\mathbf{y}\\ \mathbf{y}\left({t}_{0}\right)& ={y}_{0}\\ t& \in \left({t}_{0},T\right]\end{array}$
where $A$ is diagonalizable; $A=PD{P}^{-1}$, that the solution given by forward Euler is
$\begin{array}{r}{u}_{n}=P\left(I+hD{\right)}^{n}{P}^{-1}{u}_{0}\end{array}$
where we consider the Forward Euler method to be ${y}_{k+1}={y}_{k}+h{y}^{\prime }\left({x}_{k},{y}_{k}\right).$ N
Now, I want to show that if the real parts of ${\lambda }_{k}$ (the eigenvalues of $A$ and thus the entries of $D$) are negative, then $h<\frac{-2Re{\lambda }_{k}}{|{\lambda }_{k}{|}^{2}}$ for all $k$ implies that ${u}_{n}\to 0$.
Now, in order to show this, based on what we have obtained for the formula for ${u}_{n}$, we need only show that for each $|1+h{\lambda }_{k}|<1$. I am having a really tough time doing this, and have mostly just tried to play around with AM-GM stuff.
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beshrewd6g
You have to find the set $|1+z|<1$ in the complex plane. Squaring this inequality results in
$1+2\phantom{\rule{thinmathspace}{0ex}}Re\left(z\right)+|z{|}^{2}<1\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}Re\left(z\right)<-\frac{1}{2}|z{|}^{2}$
Now insert $z=h{\lambda }_{k}$ and that $h$ is a real number...