Re: Linear Algebra
- From: "Mkajuma" <mboghom@xxxxxxxxxxxxxx>
- Date: Sat, 26 Nov 2005 05:45:51 GMT
"Arturo Magidin" <magidin@xxxxxxxxxxxxxxxxx> wrote in message
news:dm8pq9$tbh$1@xxxxxxxxxxxxxxxxxxxxx
> In article <OPQhf.277$OO6.224@xxxxxxxxxxxxx>,
> Mkajuma <mboghom@xxxxxxxxxxxxxx> wrote:
> >My brother asked me for help with proving the following .
> >Suppose V is a complex vector space and T (is an element of ) L(V). Prove
> >that T has an invariant subspace of dimension j for each j = 1,..., dim
V.
> >
> >I have a couple of questions about this problem. Should it not be that T
is
> >a SUBSET of L(V)? Also is L a mapping from V to V?
>
> The standard notation is that
>
> L(V) = { U: V->V | U is a linear transformation }
>
> that is, the collection of all linear transformations from V to V;
> this is in fact a vector space as well, but that does not matter for
> this problem.
>
> So T being an element of L(V) means that T is a linear transformation
> from V to itself.
Got it! Thanks
>
> --
> ======================================================================
> "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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- Linear Algebra
- From: Mkajuma
- Re: Linear Algebra
- From: Arturo Magidin
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