Re: math disput btwn friends (dimension of vector spaces)
- From: "Russell" <russell@xxxxxxxx>
- Date: 29 Sep 2005 11:15:24 -0700
Russell wrote:
> SusanP wrote:
> > So because of this isomorphism, if you change
> > vector spaces between complex numbers
> > and real numbers, the dimension gets multiplied
> > by 2? (Since dimV (over complex numbers)=n,
> > then dimV(over real numbers)=2n)
>
> I guess it might depend on what you mean by "change
> vector spaces" but in the obvious meaning that comes
> to mind -- mapping (a+bi)v to the pair (av,bv) --
> yes it is always true for a finite vector space V^n.
> To prove this, you would show that n linearly
> independent vectors in the space over C map to 2n
> linearly independent vectors in the space over R.
Ick, I really botched that one, didn't I.
N vectors can't map to 2n vectors, of course.
So in fact your assertion is *NOT* true in general.
It works when your vectors are elements of C because
these elements are really pairs of real numbers in
disguise.
Shame on me for confusing vectors with their components,
ironic really when you consider the other threads I've
been posting in.
.
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