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Let $A,b$ be the coefficients of the $\nu$-stage Runge-Kutta method. On the Dahlquist test equation $y^{\prime }=\lambda y$, its stability function isBoth polynomials have degree at most $\nu$, with denominator normalized to one at zero. Order $2\nu$ implies $R(z)-e^z=O(z^{2\nu +1})$. A rational function with these degree bounds and this accuracy is uniquely the diagonal Padé approximant $[\nu /\nu {]}_{e^z}$. To see uniqueness directly, if $P/Q$ and $\tilde{P}/\tilde{Q}$ have the required accuracy, then $P\tilde{Q}-\tilde{P}Q$ is a polynomial of degree at most $2\nu$ vanishing to order at least $2\nu +1$ at zero, hence identically zero.

The diagonal exponential Padé approximant can be writtenUse the permitted Padé property that the zeros of $P_{\nu }(-z)$ lie strictly in the right half-plane and numerator and denominator are coprime. Thus $R$ has no poles in the closed left half-plane. Its real coefficients give $P_{\nu }(-iy)=\overline{P_{\nu }(iy)}$, so $|R(iy)|=1$. Also $R(z)\to (-1{)}^{\nu }$ at infinity. Applying the maximum modulus principle on left half-disks and letting their radii increase proves $|R(z)|\le 1$ for $\mathrm{Re}z\le 0$. There are no hidden stage poles there either: the irreducible Padé denominator has degree $\nu$, exhausting the possible degree of $\det (I-zA)$, so the latter cannot contain an additional canceled factor. Every such maximal-order Runge-Kutta method is A-stable. This is maximal-order Runge-Kutta methods are A-stable; it does not assert L-stability, since the stability function does not tend to zero at infinity.

Write the Runge-Kutta method coefficients as $A\in {\mathbb{R}}^{\nu \times \nu }$, $b\in {\mathbb{R}}^{\nu }$, and let $e$ be the all-ones column. On the test equation $y^{\prime }=\lambda y$, with $z=h\lambda$, the stage vector satisfies $Y=e{y}_n+zAY$ and the next value is $y_{n+1}=y_n+z{b}^{T}Y$. Hence its stability function isBoth polynomials have degree at most $\nu$, and the denominator has value one at zero. Order of a Runge-Kutta method $2\nu$ implies that this test-equation approximation agrees with $e^z$ through degree $2\nu$:By uniqueness of the diagonal Padé approximant, $R=[\nu /\nu {]}_{e^z}$. Explicitly,The permitted diagonal Padé approximant property says this rational function has no poles in the closed left half-plane and satisfies $|R(z)|\le 1$ there. ThereforeThere is also no hidden stage-solve singularity from a canceled denominator: the Padé approximant numerator and denominator are coprime and each has degree $\nu$. The original determinant denominator already has degree at most $\nu$, so, with its normalization at zero, it equals $P_{\nu }(-z)$ and cannot contain an additional canceled factor. This proves maximal-order Runge-Kutta methods are A-stable.