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 Nov11 comment If the eigenvalues are distinct then the eigenspaces are all one dimensional In your example the eigenspace for - 1 is spanned by $(1,1)$. This means that it has a basis with only one vector. It has nothing to do with the number of components of your vectors. Oct12 comment is there are specific way to solve coupled first-order differential equations with coefficients varying? I already did :-). Oct11 comment is there are specific way to solve coupled first-order differential equations with coefficients varying? And what is your particular $A$? Oct10 comment is there are specific way to solve coupled first-order differential equations with coefficients varying? I think you might find some more details in Chapter 3 of Theory of Ordinary Differential Equations by Coddington and Levinson. Oct10 comment is there are specific way to solve coupled first-order differential equations with coefficients varying? You have to distinguish between independent ($t$) and dependent ($y_1,y_2,\ldots,y_n$) variables... Oct10 comment is there are specific way to solve coupled first-order differential equations with coefficients varying? Yes, thats exactly what I was talking about. The case of commuting matrices is mentioned as Example 5.8 in the linked document. Oct9 comment is there are specific way to solve coupled first-order differential equations with coefficients varying? Yes, it was a typo. You can have a different value of the inital time, though. Oct9 awarded Commentator Oct9 revised is there are specific way to solve coupled first-order differential equations with coefficients varying? edited body Oct9 comment is there are specific way to solve coupled first-order differential equations with coefficients varying? I have improved the answer above. Hope it helps. I am not able to give you additional exact reference as I am away from my office. But I guess this should be a fairly standard stuff from any textbook on mathematical methods of quantum mechanics. I would also recommend you to look for some general theory of differential equations in Banach spaces. I will try to give you some references soon. Oct9 revised is there are specific way to solve coupled first-order differential equations with coefficients varying? added 796 characters in body; edited body; edited body; edited body Oct9 answered is there are specific way to solve coupled first-order differential equations with coefficients varying? Oct2 awarded Supporter Oct2 comment operator exponential I do not see the purpose of this. In the present case the exponential $e^{tA}$ is a linear bounded operator in $L^2(\mathbb{R})$. In general, it is defined as a solution of the initial value problem $u_t = Au$, $u(0) = u_0$, that is $(e^{tA}u_0) := u(t)$. Oct1 comment What is a countable set? Yes, you are right. One can drop "onto" property and it is fine. Oct1 answered What is a countable set? Oct1 revised Convergent subsequences added 111 characters in body Oct1 comment operator exponential In this case you have $$(e^{tA} u_0)(x) = \int_\mathbb{R} \frac{1}{\sqrt{4\pi t}} e^{-(x-y)^2/4t} u_0(y) \,\mathrm{d}y.$$ You can derive this formula employing the Fourier transform. See also "heat kernel" on wikipedia. Sep30 answered Convergent subsequences Sep30 comment operator exponential Well, in this case you have $(e^{At}u_0)(x) = u_0(x+t)$. I am assuming that $u_0$ is defined on $\mathbb{R}$. The case of a bounded interval might be complicated.