2
$\begingroup$

In the book Basic Algebra by Arthur Knapp, he states that External Direct Sum for infinitely many vector spaces can be defined as below:

$\bigoplus_{a \in A} V_a$ is the set of tuples $\{v_a\}$ of the Cartesian product $\prod_{a \in A} V_a$ with atmost finitely many $v_a$ equal to zero and vector addition and scalar multiplication defined as usual. He remarks that the basis of $\bigoplus_{a \in A}$ is the union of basis of the constituent vector spaces.

Moreover, he further defines External Direct Product for infinitely many vector spaces in the similar manner but excluding the highlighted condition.

He then remarks that unlike the External Direct Sum, the External Direct Product doesn’t have a basis which can be represented via the collective basis of the vector spaces.

I am confused about why the highlighted condition is necessary for a basis to exist. Why can’t we just use the vectors $U_{a(i)} =(0,0,...,a(i),...)$ as the basis where $a(i)$ belongs to basis of $V_a$ and $a \in A$?

$\endgroup$

1 Answer 1

2
$\begingroup$

Addition is a binary operation: it takes two vectors, and returns a vector. By induction, we can add finitely many vectors together. But we cannot add infinitely many vectors together.

So, for example, the vector $(1,1,1,1,\ldots)$ cannot be expressed as a linear combination of the vectors $\mathbf{e}_j$ (where $\mathbf{e}_j$ is an element of $\mathbb{R}^{\omega}$ in which the $i$th coordinate of $\mathbf{e}_j$ is $1$ if $i=j$ and $0$ otherwise), because by definition, a linear combination has only finitely many summands. So a linear combination of the $\mathbf{e}_i$ has only finitely many nonzero entries.

If you take the collection of vectors you propose, a linear combination of them, since by definition it can only involve finitely many vectors from your collection, will necessarily have zero entry in all but finitely many coordinates.

$\endgroup$
3
  • $\begingroup$ That makes sense. But the book says that atmost finitely many $v_a$ can be zero, which is the same as saying that that atleast infinitely many are non-zero. $\endgroup$ May 9, 2020 at 13:29
  • 1
    $\begingroup$ @ValarmathiSenthilnathan: That’s a typo (or a mistranscription). It should be “at most finitely many are nonzero.” $\endgroup$ May 9, 2020 at 16:14
  • 1
    $\begingroup$ In the second edition p.62 it says "with all but finitely many $v_\alpha$ equal to 0" $\endgroup$ Oct 7, 2020 at 10:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .