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seen Dec 22 '13 at 8:25

Nov
11
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.
Oct
12
comment is there are specific way to solve coupled first-order differential equations with coefficients varying?
I already did :-).
Oct
11
comment is there are specific way to solve coupled first-order differential equations with coefficients varying?
And what is your particular $A$?
Oct
10
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.
Oct
10
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...
Oct
10
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.
Oct
9
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.
Oct
9
awarded  Commentator
Oct
9
revised is there are specific way to solve coupled first-order differential equations with coefficients varying?
edited body
Oct
9
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.
Oct
9
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
Oct
9
answered is there are specific way to solve coupled first-order differential equations with coefficients varying?
Oct
2
awarded  Supporter
Oct
2
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 $(e^{tA}u_0) := u(t)$.
Oct
1
comment What is a countable set?
Yes, you are right. One can drop "onto" property and it is fine.
Oct
1
answered What is a countable set?
Oct
1
revised Convergent subsequences
added 111 characters in body
Oct
1
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.
Sep
30
answered Convergent subsequences
Sep
30
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.