Let $e_k$ be the standard unit sequence, where all elements but the $k$th one are zeroes, and the remaining element is $1$. Show that the union \begin{equation} \{e_k \mid k \in \mathbb N\} \cup \{(1 / k)_{k \in \mathbb N}\} \end{equation} is linearly independent on $\ell^\infty(\mathbb C)$, where the sequence space \begin{equation} \ell^\infty(\mathbb C) = \left\{ (x_k)_{k\in\mathbb N} \in \ell(\mathbb C) \mid \sup_{k\in\mathbb N} |x_k| < \infty \right\}\,. \end{equation}

An infinite set of vectors is said to be linearly independent, if the vectors of all its finite subspaces are linearly independent.

Some thoughts

Looking at the set $\{(1/k)_{k \in \mathbb N}\}$ and the standard unit sequences $e_k$ it would seem that the harmonic sequence should be expressable in therms of these unit sequences. This is because multiplying $e_k$ by $1/k$ and then summing the resulting sequences together results in the harmonic sequence.

How could I use the fact that we are operating on $\ell^\infty(\mathbb C)$ to counteract this issue?

  • 2
    $\begingroup$ The thing is that you would need an infinite sum to express the harmonic sequence as a linear expression of $e_i$. Linear dependence doesn't work over infinite sums. $\endgroup$ Jan 9 '20 at 19:24
  • $\begingroup$ I don;t know where you got the problem from, but the notation is appalling: the set comprehension on the right of the union looks nothing like the singleton set containing the sequence $\langle 1/1, 1/2, 1/3, \ldots \rangle$. $\endgroup$
    – Rob Arthan
    Jan 9 '20 at 21:13

Hint Assume that a finite linear combination of $(e_k)$ and $s=(\frac{1}{n})$ is zero. This means

$$c_1e_{k_1}+....+c_ne_{k_n}+cs=0$$ with $c$ potentially being $0$.

Look first at the "entry" $m$, where $m > k_1,.., k_n$ to deduce that $c=0$.

Next, look individually at each of the entries $k_1,...,k_n$.

  • $\begingroup$ The first entry $m$ of the linear combination, as in $c(1/k)_{k \in \mathbb N}$, or the $m$th entry of the harmonic sequence itself? The latter makes more sense. $\endgroup$
    – SeSodesa
    Jan 9 '20 at 19:51
  • 1
    $\begingroup$ @SeSodesa The entry of the linear combination. If $c_1e_{k_1}+....+c_ne_{k_n}+cs=0$ then the $m$th entry of $c_1e_{k_1}+....+c_ne_{k_n}+cs$ is zero. $\endgroup$
    – N. S.
    Jan 9 '20 at 20:22
  • 2
    $\begingroup$ The point is that once $m>k_n,$ the $m$th component of the vector sum is $c/m$ and since each component is zero, $c=0.$ And then all the other coefficients are zero because the $\{e_k\}$ are linearly independent. $\endgroup$ Jan 10 '20 at 0:54
  • $\begingroup$ Right, I got it after thinking about it a little bit. I sort of forgot, that the sequences are not finite, even if we have a finite set of them. $\endgroup$
    – SeSodesa
    Jan 10 '20 at 10:15

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