Dimension of a vector space

Question

My question is:

Is the vector space containing all periodic complex sequences a finite-dimensional vector space?

Consider $e^{(k)}$ the sequence defined by $e^{(k)}k=1$ and $e^{(k)}_j=0$ for $k\neq j$. Now take $v_p:=\sum{k=0}^{+\infty}e^{(kp)}$. These sequences are periodic. — Davide Giraudo, Oct 09 '11 at 18:12

score 3 · Answer 1 · answered Oct 10 '11 at 05:11

Davide has pretty much answered this in the comments, but here goes anyway.

Consider the sequences

$s_2=1,0,1,0,1,0,1,0,\dots$
$s_3=1,0,0,1,0,0,1,0,0,\dots$
$s_5=1,0,0,0,0,1,0,0,0,0,1,0,0,0,0\dots$
$s_7=1,0,0,0,0,0,0,1,0,0,0,0,0,0,1,\dots$

etc, where the subscripts (and the periods) are the primes. Can you convince yourself that they are linearly independent?

score 2 · Answer 2 · answered Oct 10 '11 at 05:15

Alternatively, if there were a finite basis of periodic sequences, with periods $p_1$, $\dots$, $p_n$, then every sequence would be a linear combination of the elements of that basis and, in particular, would have $q=p_1\cdots p_n$ as a period.

Since there do exist periodic sequences for which $q$ is not a period, your statement follows.

Dimension of a vector space

2 Answers2