1
$\begingroup$

In linear algebra and its application professor Gilbert Strang writes:

"A much better idea is to keep the familiar definition of length, using a sum of squares, and to include only those vectors that have a finite length:

Length squared = $||v||^2 = v_1^2 +v_2^2 + v_3^2 + ...$

The infinite series must converge to a finite sum. this leaves (1, 1/2, 1/3,...) but not (1,1,1,...). The vectors with finite length can be added $(\|v + w\| \le \|v\| + \|w\|)$ and multiplied by scalars, so they form a vector space. It is the celebrated hilbert space."

this leaves (1, 1/2, 1/3,...) but not (1,1,1,...)? does this make sense. to me partial sums of this (1,1,1,...) do not converge

  • 3
    By "this leaves" the author means "this includes." Can also think of it as in "this leaves (1, 1/2, 1/3, ...) as an element of the space but (1,1,1,...) is not an element of the space." – 2017-02-22
  • 0
    I've just noticed that Gilbert is a letter off of Hilbert. – 2017-02-22
  • 0
    It's a language thing. You start with the space of all sequences, and then _remove_ those that have infinite length. You're _left with_ elements like $(1, 1/2,1/3, \ldots)$. In other words, the process of removing the infinite elements _leaves_ $(1, 1/2,1/3, \ldots)$ behind, or it _leaves_ $(1, 1/2,1/3, \ldots)$ alone, or it _leaves_ $(1, 1/2,1/3, \ldots)$ be, or whatever else you want to use to complete that phrase. – 2017-02-22
  • 0
    @tilper : the partial sums of (1,1/2,1/3,...) diverge away to infinity as well. – 2017-02-22
  • 0
    @Arthur : the partial sums of (1,1/2,1/3,...) diverge away to infinity as well – 2017-02-22
  • 1
    But we're not looking at the partial sums. We're looking at partial sums of the _square_ of the terms in the sequence. And $1+\frac14+\frac19+\cdots$ converges just fine. – 2017-02-22
  • 0
    @arthur: what about "The infinite series must converge to a finite sum". maybe i can answer myself? he is defining sum to be the sum of squares? confirm – 2017-02-22
  • 1
    It does say $\|v\|^2 = v_1^2 +v_2^2 + v_3^2 + \cdots$ right up there in your question. You even typed it in _yourself_ a little over an hour ago (with a minor correction just now). Those are the squares of the terms of the sequence that are added together, yes. – 2017-02-22
  • 0
    @arthur: that was a typo which i corrected in the question. it is ||v||^2 = .... – 2017-02-22
  • 0
    That typo doesn't matter. It's the right-hand side that is important, and the terms there all have little $^2$'s on them. That means that you square each element before you add them together. – 2017-02-22
  • 0
    @Arthur: yes got it – 2017-02-22
  • 0
    What Arthur said. – 2017-02-22

1 Answers 1

2

What you've described is an example of a Hilbert space. Our vectors need not be lists of coordinates, and our measurement of length doesn't need to be the usual sum of squares (that is, we can define different inner products). However: you can indeed say that $(1,1,\dots)$ fails to be an element of the Hilbert space that you've described.