Find the Kernel & Image of A: $\mathbb{R}^\infty \rightarrow \mathbb{R}^\infty$

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Homework Statement


Find the kernel and image of the linear function A: \mathbb{R}^\infty \rightarrow \mathbb{R}^\infty defined on the vector space (with usual operations) of sequences of real numbers x \in \mathbb{R}^\infty, x = (x_1, x_2,...). given by A(x) = (y_1, y_2, ...) with y_k = x_{2k+1} - 2x_k.

Homework Equations



Ker(A) = \{ x \in \mathbb{R}^\infty : Ax = (0, 0, 0,...) \}
and the standard notation for the image of A

The Attempt at a Solution


For the kernel I tried to check formulas to solve finite difference equations, since I need to solve x_{2k+1} - 2x_k =0 but I can only find the usual method when we have k+1, k+2 or k+r terms involved, not when one of the terms is of the form x_{ak+b} when a is not 1.

I did it by hand, plugging in values and I see a pattern, and I know that after a while i could come up with a general form for a basis of the (as I can see) infinite dimensional Kernel. I would like to solve it more decently.

For the second part (the image of A) I am still lost, since I was hoping to get a general solution and from there see which sequences are not allowed, somehow... Does anyone know how to solve this recurrence equation?
Or at least how to solve this problem, if the solving of the equation is not absolutely necessary? (but I would love to learn how to solve these equations).
Thanks!
 
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hi arestes! :smile:

(try using the X2 icon just above the Reply box :wink:)

the key is to find the pattern for the indices:

k, 2k+1, 2(2k+1) + 1 etc :smile:
 
Hi Tiny Tim, I was doing what you said for the kernel. Any ideas for the image?
thanks! :D
 
hi arestes! :wink:
arestes said:
Hi Tiny Tim, I was doing what you said for the kernel. Any ideas for the image?
thanks! :D

ooh, i didn't get that far :redface:

i sort of assumed that if you managed to find the kernel, the image would be easy …

would you like to show us how far you've got? :smile:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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