Re: Direct Sums

In article <2194777.1128013672052.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
SportsNut <dsfjk@xxxxxxxxx> wrote:
>>From the definition of direct sums:
>A vector space is called a direct sum of
>W1 and W2 if W1 and W2 are subspaces of V
>such that W1*intersect*W2={0} AND
>W1+W2=V. Direct sum is denoted
>V=W1 *direct sum* W2.
>
>My question is:
>If i only know that V=R(T)+N(T),
>can I prove that V=R(T) *direct sum* N(T)?
>Or do I NEED the fact that
>R(T)*intersection*N(T)={0}?

There are many ways to establish that the intersection must be zero
without exactly calculating it, certainly. Formally, you need to
establish that the intersection is trivial somehow, or apply a theorem
that you know guarantees this fact.

>NOTES: R(T)= range of T
>N(T)= null space of T
>V= finite dimensional vector space
>T:V->V is linear

So, for example, if you know the Dimension Theorem aka the
Rank-Nullity Theorem, then you know that dim(V) =
dim(R(T))+dim(N(T)). If you happen to know as well that

dim(W1+W2) = dim(W1) + dim(W2) - dim(W1 intersect W2)

then you would be able to deduce from V=R(T)+N(T) and the Dimension
Theorem that dim(R(T) intersect N(T))=0, from which it would follow
that R(T) intersect N(T) = {0}, from which it would follow that V is
the direct sum of R(T) and N(T).



--
======================================================================
"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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