[seqfan] Re: Guess the Triangle

[Ron Hardin]

Must be triangular (proof)

Move the zero diagonal basis to the top.

Read column j as saying where a post-multiplied vector element j gets 
distributed to.

For j>0 elements get distributed to themselves and at most to 0, since only the 
0 element gets annihilated on iteration, and so allows the matrix to be 
idempotent when iterated.

For j=0 elements may be distributed anywhere but 0, so that the rank is n-1 and 
since the 0 element does not remain at itself and so does not iterate.

If both the top row and left column are nonzero, then there's a j->0->k path 
that distributes an element from j to 0 to k at the second iteration, j>0  and 
k>0, and the matrix is not idempotent.

(Argument would not work if negative elements were permitted)

 rhhardin at mindspring.com
rhhardin at att.net (either)



----- Original Message ----
> From: Rob Pratt <Rob.Pratt at sas.com>

> Sorry, triangular only if the 0 appears first or last on the diagonal.  The  
>correct characterization seems to be: differs from the identity matrix in  
>exactly one or row or column, with 0 as the diagonal element of that  column.

...


> >> been counted twice, so subtract n.  It  remains  to show that such matrices 
>must 
>
> >> be triangular so that  these are the only  possibilities.