a matrix transformation

F. Jooste pin at myway.com
Thu Dec 16 02:01:58 CET 2004


Seqfans, happy early holiday season to anyone who is celebrating.

I am wondering if the following observation is worth noting in the OEIS, and exactly where the note should be made if so warranted.

Consider a square matrix of all positive real numbers, such as this 2 by 2 one,

A=
a1 & a2 \\ 
b1 & b2.

Let's say we want to transform this to get alternating terms negative (like a checkered board), the upper left term remains positive, like this,

B=
a1 & -a2 \\ 
-b1 & b2.

In order to do this, I want to define a matrix function (skewed, I am not sure what else to call it; it's a diagonal of +/-1 and zero everywhere else),

S(x1,x2)=
0 & (-1)^x1 \\ 
(-1)^x2 & 0,

where x1 and x2 are chosen from the set {0,1}.  More generally,

S(x1,x2,...,xn)=
0 & 0 & ... & 0 & (-1)^x1 \\ 
0 & 0 & ... & (-1)^x2 & 0 \\ 
................. \\
(-1)^xn & 0 & ... & 0 & 0.

Then, back to n=2, the transformation looks like this:

B = S(w1,w2) * S(x1,x2) * A * S(y1,y2) * S(z1,z2).

Now to the point:  If we look at the total number of choices for {w1,w2,x1,x2,y1,y2,z1,z2} that successfully transform A into B, then it seems to me to be 2^5 in this case.  For a square matrix A of size n by n, we want to count the choices for
{w1,w2,...,wn,x1,x2,...,xn,y1,y2,...,yn,z1,z2,...,zn}, which suggests the following sequence.

Let a(n) be the total number of choices from {0,1} to transform A into B, where n is the size of the square matrix A (offset n=1).  Then

{a(n)} = {2^3, 2^5, 2^7, ...} = {2^(2*n+1)}.

I have tested this from n=1 to n=3, but haven't actually proven it.

A004171 is the sequence 2^(2n+1).  Is this observation worth putting into a note there?  Is it too complicated or arbitrary to bother? 

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