Re: onto equivalents
- From: magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin)
- Date: Tue, 29 Nov 2005 17:04:41 +0000 (UTC)
In article <438c7893$1_1@xxxxxxxxxxxxxxxxxxx>,
Gary Weselle <weselle_g@xxxxxxxxxxx> wrote:
>Hello
>
>I would appreciate some help with explaining why the statements below are
>equivalents "proofs":
>
>
>
>Let A: R^n -> R^m
Then the following are equivalent:
>A is onto
>
>The columns of A spans R^m
>
>
>
>My understanding of onto is, given f(x) = y, for each y there exists one or
>more x.
which is a solution to the equation f(x)=y.
>That is to say, there exists one or more solution(s), that is to say, one or
>more linear combinations of the independent columns of A, i.e, the columns
>of A spans the R^m?
If the columns of A span R^m, that means that for every vector b in R^m,
there is some linear combination of the columns of A which is equal to
b. Now, every product the form Ax, with x a vector in R^n, corresponds
to a linear combination of the columns. That is:
Ax = b has a solution
is the same as
there is a linear combination of the columns of A which is equal to b.
So:
Columns of A span R^m
is equivalent to
For every b in R^m, there is a linear combination of the columns of
A which is equal to b
which is equivalent to
For every b in R^m, there is an x in R^n such that Ax = b
which is equivalent to
Thinking of A as a function, A:R^n --> R^m, for every b in R^m, there is
an x in the domain of A such that A(x)=b.
which is equivalent to
Thinking of A as a function, A:R^n --> R^m, for every b in R^m,
given the equation A(x)=b, there always exists an x that is a
solution to the equation
which is equivalent to
Thinking of A as a map, A:R^n --> R^m, A is onto.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx
.
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