It is straightforward to see that if the path is constant then hyperbolicity is the equivalent to the invertibility of the operator for example via the Fourier transform (see Atiyah Patodi Singer (Spectral Asymmetry and Riemann Geometry I) for example).
It is then straightforward to see that if the $A(t)$ is asymptotically hyperbolic then the operator is Fredholm by a parametrix patching argument.
There is then a well-defined index problem. The index is then equal to the spectral flow of the family $A(t)$, namely the intersection number of the spectrum with the imaginary axis. This can be easily proved using the homotopy invariance of the index and reducing to the case where the $A(t)$ are simultaneously diagonalizable with diagonal entries being either constant or $\pm\tanh(t)$. This can be solved explicitly with kernel or cokernel $\text{sech}(t)$ depending on the sign.

Back to your problem, the sign of the index provides an obstruction to the operator being injective or subject but not a complete one. For example if $A(t)$ is direct sum of two matrix paths $A_-(t)$ and $A_+(t)$ of spectral flow $-n$ and $n$ respectively, then the corresponding operator has kernel and cokernel of dimension at least $n$ but the total operator has index $0$. On the other hand, in the one-dimensional case it seems straightforward to see that it is a complete obstruction. Writing $\lambda(t)=a_{11}(t)$ the hyperbolicity assumption means that $a(t)$ converges at $\pm\infty$ to constants $a_\pm$ of the same sign. An element of the kernel has the form: $$ f(t)=c+\exp[-\int_0^t \lambda(s)ds] $$ Regardless of how you choose the constant the operator is exponentially growing at one end. A similar argument applies to the cokernel.

It is straightforward to see that if the path is constant then hyperbolicity is the equivalent to the invertibility of the operator for example via the Fourier transform (see Atiyah Patodi Singer (Spectral Asymmetry and Riemann Geometry I) for example).
It is then straightforward to see that if the $A(t)$ is asymptotically hyperbolic then the operator is Fredholm by a parametrix patching argument.
There is then a well-defined index problem. The index is then equal to the spectral flow of the family $A(t)$, namely the intersection number of the spectrum with the imaginary axis. This can be easily proved using the homotopy invariance of the index and reducing to the case where the $A(t)$ are simultaneously diagonalizable with diagonal entries being either constant or $\pm\tanh(t)$. This can be solved explicitly with kernel or cokernel $\text{sech}(t)$ depending on the sign.

Back to your problem, the sign of the index provides an obstruction to the operator being injective or subject but not a complete one. For example if $A(t)$ is direct sum of two matrix paths $A_-(t)$ and $A_+(t)$ of spectral flow $-n$ and $n$ respectively, then the corresponding operator has kernel and cokernel of dimension at least $n$ but the total operator has index $0$. On the other hand, in the one-dimensional case it seems straightforward to see that it is a complete obstruction. Writing $\lambda(t)=a_{11}(t)$ the hyperbolicity assumption means that $a(t)$ converges at $\pm\infty$ to constants $a_\pm$ of the same sign. An element of the kernel has the form: $$ f(t)=c+\exp[-\int_0^t \lambda(s)ds] $$ Regardless of how you choose the constant, $f(t)$ is exponentially growing at one end or the other. A similar argument applies to the cokernel.

It is straightforward to see that if the path is constant then hyperbolicity is the equivalent to the invertibity of the operator for example via the Fourier transform (see Atiyah Patodi Singer (Spectral Asymmetry and Riemann Geometry I for example.)
It is then straight forward to see that if the $A(t)$ is asymptotically hyperbolic then the operator is Fredholm by a parametrix patching argument.
There is then a well defined index problem. The index is then equal to the spectral flow of the family $A(t)$, namely the intersection number of the spectrum with the imaginary axis. Then can be easily proved using the homotopy invariance of the index and reducing to the case where the $A(t)$ are simultaneously diagonalizable with diagonal entries being either constant or $\pm\tanh(t)$. This can be solved explicitly with kernel or cokernel $\sech(t)$ depending on the sign.

Back to your problem, the sign of the index provides an obstruction to the operator being injective or subject but not a complete one. For example if $A(t)$ is direct sum of two matrix paths $A_-(t)$ and $A_+(t)$ of spectral flow $-n$ and $n$ respectively, then the correspond operator has kernel and cokernel of dimension at least $n$ but the total operator has index $0$. On the other hand in the one dimensional case it seems straight forward to see that it is a complete obstruction. Writing $\lambda(t)=a_{11}(t)$ the hyperbolicity assumption means that $a(t)$ converges at $\pm\infty$ to constants $a_\pm$ of the same sign. An element of the kernel has the form: $$ f(t)=c+\exp[-\int_0^t \lambda(s)ds] $$ regardless of how you choose the constant the operator is exponential growing at one end. As similar argument applies to the cokernel.

It is straightforward to see that if the path is constant then hyperbolicity is the equivalent to the invertibility of the operator for example via the Fourier transform (see Atiyah Patodi Singer (Spectral Asymmetry and Riemann Geometry I) for example).
It is then straightforward to see that if the $A(t)$ is asymptotically hyperbolic then the operator is Fredholm by a parametrix patching argument.
There is then a well-defined index problem. The index is then equal to the spectral flow of the family $A(t)$, namely the intersection number of the spectrum with the imaginary axis. This can be easily proved using the homotopy invariance of the index and reducing to the case where the $A(t)$ are simultaneously diagonalizable with diagonal entries being either constant or $\pm\tanh(t)$. This can be solved explicitly with kernel or cokernel $\text{sech}(t)$ depending on the sign.

Back to your problem, the sign of the index provides an obstruction to the operator being injective or subject but not a complete one. For example if $A(t)$ is direct sum of two matrix paths $A_-(t)$ and $A_+(t)$ of spectral flow $-n$ and $n$ respectively, then the corresponding operator has kernel and cokernel of dimension at least $n$ but the total operator has index $0$. On the other hand, in the one-dimensional case it seems straightforward to see that it is a complete obstruction. Writing $\lambda(t)=a_{11}(t)$ the hyperbolicity assumption means that $a(t)$ converges at $\pm\infty$ to constants $a_\pm$ of the same sign. An element of the kernel has the form: $$ f(t)=c+\exp[-\int_0^t \lambda(s)ds] $$ Regardless of how you choose the constant the operator is exponentially growing at one end. A similar argument applies to the cokernel.