1
$\begingroup$

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?

$\endgroup$
2
  • 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
2
$\begingroup$

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 $k_1,...,k_n$.

$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.