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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