# Are geometric progressions forming a vector space?

:) I'm new here and I have a question for which I didn't find any hint on the website, yet. Maybe you can help me :).

I am trying to find out whether geometric progressions with real entries are R-vector spaces.

I found the solution for an arithmetic progression, but now for geometric ones I struggle to see it intuitively, especially on the closure under addition statement. I.E is the sum of two geometric progressions also a GP ?

$a_{n+1}=r\cdot a_{n}$ , $b_{n+1}=r\cdot b_{n} \\$

$(a+b)_{n+1}=(a+b)_{n}\cdot r$

Thank you very much in advance :)

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Hint: You could just try your favorite two GPs: mine are $1,2,4,8,16,\ldots$ and $1,3,9,27,81 \ldots$. The sum of these is $2,5,13,35,97 \ldots$ Now calculate the ratio between neighboring terms.

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As your computation shows, the sum of two geometric progressions with the same common ratio $r$ is a geometric progression with common ratio $r$. For that, we have to think of the sequence $0,0,0,\dots$ as a geometric progression with common ratio $r$ for every $r$, but that's OK.

It is not hard to verify that for any fixed $r$, the collection of geometric sequences with common ratio $r$, under the obvious addition and multiplication by constants, is a vector space over the reals.

In general, the sum of two geometric progressions with different common ratio is not a geometric progression. So the collection of all geometric progressions, under the usual sum, is not a vector space.

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Follow up question: does the second paragraph imply that this vector space has dimension $1,$ and that $(1, r, r^2, \ldots)$ is a basis vector? – user2468 Mar 20 '12 at 20:28
@J.D.: Yes, it is a pretty boring vector space. – André Nicolas Mar 20 '12 at 20:37