Consider the second-order ordinary differential system $$ A y'' + i B y' - C y = 0, $$ where $A$ and $C$ are real-valued $d\times d$ SPD matrices satisfying $$ I \le A,C \le 2I, $$ and $B$ is real-valued and symmetric, having operator 2-norm less than 2 (and $i = \sqrt{-1}$). I would like to show that
EDIT 1: I think one needs to assume that $\|B\|<2$, since otherwise as @Iosif Pinelis example in the comments shows, the eigenvalues may be purely imaginary.
EDIT 2: I changed the notation so my answer become clearer. In the original answer, I had the following fact in mind: if $P(\xi)$ is the principle symbol of a linear $2$-differential operator satisfying $P(\xi)\geq c|\xi|^{2}$, then the eigenvalues of the $2$-order linear ODE $P(\xi+i\frac{d}{dt}\nu)y(t)=0$ must have the desired properties in the question. I basically reconstructed such a symbol from the given ODE, but then I reckoned this will not be needed in the actual answer.
Under the assumption that $\|B\|<2$, let us see that all eigenvalues have either positive or negative real part. Consider the matrix-valued polynomial of degree 2 in $\lambda$, $P:\mathbb{R}\rightarrow\mathrm{End}{\mathbb(\mathbb{R}^{d})}$, given by $$P(\lambda)=\lambda^{2}A-\lambda B+C$$ Note how for every $\lambda\in\mathbb{R}$, by the assumptions in the question we find $$c(1+\lambda^{2})I\leq P(\lambda)$$ For some constant $c>0$. In particular, $P(\lambda)$ is nonsingular for all $\lambda\in\mathbb{R}$.
By formally allowing $\lambda$ to be complex, we note how the eigenvalue problem for the ODE in the question becomes for $y(t)=y_{0}e^{\lambda t}$, $$P(i\lambda)y_{0}=0$$ For $\lambda=i\eta$ with $\eta\in\mathbb{R}$, this reads, $$P(-\eta)y_{0}=0$$ which implies $y_{0}=0$. Therefore, the eigenvalues of the ODE can't be purely imaginary, and must contain a real part.
This implies that the fundamental solutions exponentially decreases when either $t\rightarrow\infty$ or $t\rightarrow-\infty$. Let $M^{+}$ denote the space of $t\rightarrow\infty$ exponentially decreasing solutions and by $M^{-}$ the space of $t\rightarrow-\infty$ exponentially decreasing solutions. Let $I:M^{\pm}\rightarrow \mathbb{R}^{d}$ be the map $Iy=y(0)$. This map is clearly linear. We prove that this map is injective, hence $\operatorname{dim}{M^{\pm}}\leq d$. Since the space of solutions is $2d$ dimensional, this implies that $\operatorname{dim}{M^{\pm}}=d$, which is the required result. The proof also shows that all fundamental solutions must be of the form $y(t)=y_{0}e^{\lambda t}$, because if $y(t)=y_{0}t^{k}e^{\lambda t}$ for some $k\ne 0$, then $y(0)=0$, hence $y=0$ identically.
Without loss of generality, let $y\in M^{+}$ with $y(0)=0$. Denote by $\hat{y}(\eta)$ the fourier transform of $y$, albeit extneded to satisfy $y(t)=0$ for $t\leq 0$. Since $P(\lambda)$ is positive definite for all $\lambda\in\mathbb{R}$, we may write: $P(\eta)=M(\eta)M(\eta)$, where $M(\eta)$ is a matrix-valued polynomial of degree at most 1 in $\eta$.
By formally replacing $\eta$ with an ordinary derivative $i\frac{d}{dt}$,
the ODE can formally be written as,
$$P(i\frac{d}{dt})y(t)=0$$
Since $y(0)=0$ and the solution is rapidly decreasing, we can integrate by parts the following equation to obtain,
$$0=(P(i\frac{d}{dt})y,y)_{L^{2}(\mathbb{R^+})}=(M(i\frac{d}{dt})y,M(i\frac{d}{dt})y)_{L^{2}(\mathbb{R^+})}$$
By the Plancheral theorem, this quantity becomes by applying the Fourier transform on each flank,
$$\int_{-\infty}^{\infty}(P(\eta)\hat{y}(\eta),\hat{y}(\eta))_{\mathbb{R}^{d}}d\eta=0$$.
And since $P(\eta)\geq c(1+\eta^{2})I$ for all $\eta\in\mathbb{R}$, this yields $\hat{y}(\eta)=0$, which is to say $y=0$. This proves that $I$ was injective, and the solution space decomposes as described.
