[proofplan] The proof is componentwise. For each diagonal entry $\lambda_j$, positivity of the step size implies that $k\lambda_j$ remains in the closed left half-plane. A-stability therefore bounds the corresponding scalar amplification factor by one. The assumed diagonal decoupling gives a scalar recurrence on each component, and induction turns that recurrence into an explicit formula whose absolute value is non-increasing in $n$. [/proofplan]

[step:Place each scaled diagonal eigenvalue in the A-stable region] Fix an index $j\in\{1,\dots,m\}$. Define the scaled scalar \begin{align*} z_j:=k\lambda_j\in\mathbb{C}. \end{align*} Since $k>0$ is real and $\lambda_j\in\mathbb{C}_{-}$, we have \begin{align*} \operatorname{Re}(z_j)=\operatorname{Re}(k\lambda_j)=k\operatorname{Re}(\lambda_j)\le 0. \end{align*} Thus $z_j\in\mathbb{C}_{-}$. Since the method is A-stable, $\mathbb{C}_{-}\subset U$, so $z_j\in U$, and \begin{align*} |R(z_j)|\le 1. \end{align*} Equivalently, \begin{align*} |R(k\lambda_j)|\le 1. \end{align*} [/step]

[step:Use diagonal decoupling to obtain the scalar recurrence] By the assumed componentwise decoupling property for diagonal linear systems, applying the method to $y'(t)=Ay(t)$ with step size $k$ gives, for every $n\in\mathbb{N}\cup\{0\}$ and every $1\le j\le m$, \begin{align*} Y_{n+1,j}=R(k\lambda_j)Y_{n,j}. \end{align*} This is exactly the asserted componentwise update formula. [/step]

[step:Iterate the scalar recurrence and take absolute values]Fix $j\in\{1,\dots,m\}$, and define the scalar amplification factor \begin{align*} a_j:=R(k\lambda_j)\in\mathbb{C}. \end{align*} The recurrence from the previous step is \begin{align*} Y_{n+1,j}=a_jY_{n,j}. \end{align*} We prove by induction on $n\in\mathbb{N}\cup\{0\}$ that \begin{align*} Y_{n,j}=a_j^nY_{0,j}. \end{align*} For $n=0$, this is $Y_{0,j}=a_j^0Y_{0,j}$, since $a_j^0=1$. If the identity holds for some $n\in\mathbb{N}\cup\{0\}$, then the recurrence gives \begin{align*} Y_{n+1,j}=a_jY_{n,j}=a_j(a_j^nY_{0,j})=a_j^{n+1}Y_{0,j}. \end{align*} Thus the formula holds for every $n\in\mathbb{N}\cup\{0\}$. Taking absolute values and using multiplicativity of the complex modulus gives \begin{align*} |Y_{n,j}|=|a_j|^n|Y_{0,j}|. \end{align*} From the A-stability estimate in the first step, $|a_j|=|R(k\lambda_j)|\le 1$. Hence $|a_j|^n\le 1$ for every $n\in\mathbb{N}\cup\{0\}$, and therefore \begin{align*} |Y_{n,j}|\le |Y_{0,j}|. \end{align*} Since $j$ was arbitrary, the decay estimate holds for every component.[/step]

[guided]Fix a component index $j\in\{1,\dots,m\}$. The point of working componentwise is that the diagonal decoupling assumption has reduced the vector method to a scalar recurrence. Define the scalar amplification factor \begin{align*} a_j:=R(k\lambda_j)\in\mathbb{C}. \end{align*} For this fixed component, the recurrence is \begin{align*} Y_{n+1,j}=a_jY_{n,j} \end{align*} for every $n\in\mathbb{N}\cup\{0\}$. We first solve this recurrence exactly. We claim that \begin{align*} Y_{n,j}=a_j^nY_{0,j} \end{align*} for every $n\in\mathbb{N}\cup\{0\}$. The base case is $n=0$: by the convention $a_j^0=1$, the identity reads \begin{align*} Y_{0,j}=a_j^0Y_{0,j}. \end{align*} Now assume the identity holds at some time index $n\in\mathbb{N}\cup\{0\}$. Substituting this inductive hypothesis into the recurrence gives \begin{align*} Y_{n+1,j}=a_jY_{n,j}=a_j(a_j^nY_{0,j})=a_j^{n+1}Y_{0,j}. \end{align*} This proves the explicit formula by induction. Now we convert the explicit formula into a decay estimate. Taking absolute values and using the multiplicative property $|uv|=|u||v|$ for complex numbers gives \begin{align*} |Y_{n,j}|=|a_j^nY_{0,j}|=|a_j|^n|Y_{0,j}|. \end{align*} Because $k>0$ is real and $\lambda_j\in\mathbb{C}_{-}$, the scaled value $k\lambda_j$ satisfies \begin{align*} \operatorname{Re}(k\lambda_j)=k\operatorname{Re}(\lambda_j)\le 0. \end{align*} Hence $k\lambda_j\in\mathbb{C}_{-}$. A-stability says that $\mathbb{C}_{-}\subset U$ and $|R(z)|\le 1$ for every $z\in\mathbb{C}_{-}$, so $R(k\lambda_j)$ is defined and \begin{align*} |a_j|=|R(k\lambda_j)|\le 1. \end{align*} Therefore $|a_j|^n\le 1$ for every $n\in\mathbb{N}\cup\{0\}$. Substituting this into the previous identity yields \begin{align*} |Y_{n,j}|\le |Y_{0,j}|. \end{align*} Because the index $j$ was arbitrary, the same argument applies to every diagonal component.[/guided]