Understanding Direct Products of Vector Spaces: Cooperstein's Example 1.17

[Math Amateur]

In Bruce Cooperstein's book: Advanced Linear Algebra, he gives the following example on page 12 in his chapter on vector spaces (Chapter 1) ... ...View attachment 4886I am finding it difficult to fully understand this example ... ...

Can someone give an example using Cooperstein's construction ... using, for clarity, his notation ... ?

If we take $$I = \{ 1, 2, \ ... \ ... \ , n \}$$ ... ... then I am used to thinking that the direct product of vector spaces $$U_1, U_2, \ ... \ ... \ U_n$$ is the set of all n-tuples $$( u_1, u_2, \ ... \ ... \ u_n )$$ with addition and scalar multiplication defined componentwise ... ...

BUT ... how do we square this with Cooperstein's definition/construction of the direct product of a set of vector spaces ... ...

Hope someone can help clarify the above issues ... ...

Peter

 

[Deveno]

Consider an $n$-tuple in $\Bbb R^n$, say $u = (x_1,\dots,x_n)$.

We can regard this as a function $f:\{1,\dots,n\}\to \Bbb R$, namely, the function given by:

$f(i) = x_i$.

So, in Cooperstein's notation, if we call the set $\{1,\dots,n\}$ say, $I$, we have:

$\Bbb R^n = \Pi_{i \in I}\ \Bbb R = \Bbb R\oplus \cdots \oplus \Bbb R$ (an $n$-fold direct product).

If we have $v = (y_1,\dots,y_n)$, we can similarly define $v$ as the function $g:I \to \Bbb R$ given by:

$g(i) = y_i$.

Normally, we think of $u+v$ as being defined as "component-wise" addition, that is:

$u + v = (x_1 + y_1,\dots,x_n + y_n)$.

If we define $f+g$ by $(f+g)(i) = f(i) + g(i)$ as Cooperstein does, we get that $f+g$ maps $i$ to $x_i + y_i$. So it accomplishes the same thing.

The advantage to Cooperstein's definition, is that if $I$ is not a FINITE set, we can still define the direct product of a number of spaces corresponding to the infinite number of things in $I$ ($I$ is called the *indexing* set).

Convince yourself that if $I = \Bbb N$, and $U_i = F$ for all $i \in I$, that the resulting space we get is isomorphic as a $F$-vector space to $F[x]$. It is common to refer to the image $f(i)$ as the "$i$-th coordinate" of a vector when all the $U_i$ coincide with the underlying field, and as the "$i$-th component" when the $U_i$ are subspaces of the product.

 

[Math Amateur]

Consider an $n$-tuple in $\Bbb R^n$, say $u = (x_1,\dots,x_n)$.

We can regard this as a function $f:\{1,\dots,n\}\to \Bbb R$, namely, the function given by:

$f(i) = x_i$.

So, in Cooperstein's notation, if we call the set $\{1,\dots,n\}$ say, $I$, we have:

$\Bbb R^n = \Pi_{i \in I}\ \Bbb R = \Bbb R\oplus \cdots \oplus \Bbb R$ (an $n$-fold direct product).

If we have $v = (y_1,\dots,y_n)$, we can similarly define $v$ as the function $g:I \to \Bbb R$ given by:

$g(i) = y_i$.

Normally, we think of $u+v$ as being defined as "component-wise" addition, that is:

$u + v = (x_1 + y_1,\dots,x_n + y_n)$.

If we define $f+g$ by $(f+g)(i) = f(i) + g(i)$ as Cooperstein does, we get that $f+g$ maps $i$ to $x_i + y_i$. So it accomplishes the same thing.

The advantage to Cooperstein's definition, is that if $I$ is not a FINITE set, we can still define the direct product of a number of spaces corresponding to the infinite number of things in $I$ ($I$ is called the *indexing* set).

Convince yourself that if $I = \Bbb N$, and $U_i = F$ for all $i \in I$, that the resulting space we get is isomorphic as a $F$-vector space to $F[x]$. It is common to refer to the image $f(i)$ as the "$i$-th coordinate" of a vector when all the $U_i$ coincide with the underlying field, and as the "$i$-th component" when the $U_i$ are subspaces of the product.

Hi Deveno ... thanks for the help ...

Until I received your post I was having real trouble seeing how a function $$f$$ was identical to (or could be equivalent to) an $$n$$-tuple or an indexed set of values ... then after reading through your post I realized that the set of function values, $$f(i)$$ is, of course, indexed by the set $$I$$ in the same way as an $$n$$-tuple ... and so the function $$f$$ gives essentially the same information as an $$n$$-tuple ...Just a further question, however ... ... ... you wrote:

"Convince yourself that if $I = \Bbb N$, and $U_i = F$ for all $i \in I$, that the resulting space we get is isomorphic as a $F$-vector space to $F[x]$. ... ... "

I am having trouble seeing this ... can you help further ... it looks a really interesting point ...

Hope you can help ...

Peter

 

[Deveno]

To any polynomial $f(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n$ we can assign the $n$-tuple:

$(a_0,a_1,a_2,\dots,a_n)$ (our indexing set is infinite, because $n$ might be arbitrarily large).

Your job is to show that if we call this $n$-tuple $[f]$, that the mapping:

$f(x) \mapsto [f]$ is $F$-linear.

 

[Math Amateur]

To any polynomial $f(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n$ we can assign the $n$-tuple:

$(a_0,a_1,a_2,\dots,a_n)$ (our indexing set is infinite, because $n$ might be arbitrarily large).

Your job is to show that if we call this $n$-tuple $[f]$, that the mapping:

$f(x) \mapsto [f]$ is $F$-linear.

Thanks Deveno ...

Can you clarify exactly what is meant by F-linear ...

Peter

 

[Deveno]

By definition an $F$-linear map is an $F$-module homomorphism, in other words it preserves the module addition, and the action of $F$ upon the underlying $F$-module $V$. This is often written:

$T(\alpha u + \beta v) = \alpha T(u) + \beta T(v)$.

For example, on the vector space of real polynomials, $\Bbb R[x]$, the map $D$ given by $D(f(x)) = f'(x)$

(here $f'(x)$ is the *formal derivative of $f$*, if:
$f(x) = a_0 + a_1x +\cdots + a_nx^n$, then $f'(x) = a_1 + 2a_2x +\cdots + na_nx^{n-1}$)

is a $\Bbb R$-linear map, since:

$D(f(x) + g(x)) = (f+g)'(x) = f'(x) + g'(x) = D(f(x)) + D(g(x))$

and:

$D(\alpha f(x)) = (\alpha f)'(x) = \alpha(f'(x)) = \alpha D(f(x))$.