(I would like to personally thank Terry Tao for his very helpful blogs on random matrices -- see here. A closer look at his ideas there helped me with my research on OPNs.)
Consider the equation
A + B = C + D (where, for now, we assume all of A, B and C are positive, D is unrestricted)
Assuming a nonzero sum C + D:
(A + B)/(C + D) = 1 (from which it also follows that the sum A + B is nonzero)
Assuming C is not equal to D:
[(A + B)(C - D)]/[(C + D)(C - D)] = 1
[ (AC - BD) + (BC - AD) ] / [ C^2 - D^2 ] = 1
[ (AC - BD)/(C^2 - D^2) ] + [ (BC -AD)/(C^2 - D^2) ] = 1
Passing on to determinant notation:
det X + det Y = det Z
where the 2x2 matrices X, Y and Z are defined as follows:
X
A B
D C
Y
B A
D C
Z
C D
D C
(The numbers A, B, C and D were laid out in their respective positions by "basket-weaving".)
Some linear algebra:
The traces of these matrices are:
tr(X) = A + C
tr(Y) = B + C
tr(Z) = 2C
Solving for the eigenvalues of each of these 3 matrices:
det(X - xI) = 0 = (A - x)(C - x) - BD = x^2 - (A + C)x + (AC - BD)
det(Y - yI) = 0 = (B - y)(C - y) - AD = y^2 - (B + C)y + (BC - AD)
det(Z - zI) = 0 = (C - z)(C - z) - D^2 = z^2 - (2C)z + (C^2 - D^2)
It is evident that the characteristic polynomial P(m) for a 2x2 matrix M then takes the form:
P(m) = m^2 - [tr(M)]m + [det(M)]
and its discriminant d(P(m)) is then:
d(P(m)) = [tr(M)]^2 - 4[det(M)]
In particular, if s and t are the eigenvalues of a particular 2x2 matrix M, then:
det(M) = st
tr(M) = s + t
Consequently (assuming s <= t):
s = (1/2)*([tr(M)] - SQRT[d(P(m)])
t = (1/2)*([tr(M)] + SQRT[d(P(m)])
For the matrices X, Y and Z above, the respective discriminants are:
d(P(x)) = [tr(X)]^2 - 4[det(X)] = (A + C)^2 - 4(AC - BD) = (A^2 + 2AC + C^2) - 4AC + 4BD
= (A - C)^2 + 4BD
d(P(y)) = [tr(Y)]^2 - 4[det(Y)] = (B + C)^2 - 4(BC - AD) = (B^2 + 2BC + C^2) - 4BC + 4AD
= (B - C)^2 + 4AD
d(P(z)) = [tr(Z)]^2 - 4[det(Z)] = (2C)^2 - 4(C^2 - D^2) = 4C^2 - 4C^2 + 4D^2 = (2D)^2
(Note that, if we want all three discriminants to be perfect squares, we can take D = 0. The original equation under consideration then becomes A + B = C. Indeed, I had the ABC conjecture in mind while trying to do these things the other day -- but it turned out these were closely related to another open problem.)
If the discriminant d(P(m)) is a (rational) square, it follows that the eigenvalues s and t are also rational, and conversely.
We now compute the eigenvalues (under the assumption s <= t):
s_X = (1/2)*(A + C - SQRT((A - C)^2 + 4BD))
t_X = (1/2)*(A + C + SQRT((A - C)^2 + 4BD))
s_Y = (1/2)*(B + C - SQRT((B - C)^2 + 4AD))
t_Y = (1/2)*(B + C + SQRT((B - C)^2 + 4AD))
s_Z = (1/2)*(2C - SQRT((2D)^2)) = (1/2)*(2C - 2D) = C - D
t_Z = (1/2)*(2C + SQRT((2D)^2)) = (1/2)*(2C + 2D) = C + D
Under the additional assumption that D = 0, these expressions simplify to:
s_X = A
t_X = C
s_Y = B
t_Y = C
s_Z = C
t_Z = C
With this assumption about the vanishing of D, the original matrices X, Y and Z become:
X'
A B
0 C
Y'
B A
0 C
Z'
C 0
0 C
Thus, we have two upper-triangular matrices X' and Y', and Z' which is a scalar multiple of the 2x2 identity matrix. But that is not really the crucial point.
Note that we have shown that:
A = s_X <= t_X = C
B = s_Y <= t_Y = C
C = s_Z <= t_Z = C
without knowing the actual values of A, B and C. How did this happen?
Hint: SQRT(x^2) = ABS(x), where ABS is the absolute value function.
In summary: Note how the value of D (which was initially unrestricted, and which we later take to be zero for purposes of "simplifying" the computations for the eigenvalues) imposed A LOT of STRUCTURAL RESTRICTIONS on the three matrices.
Later during the day, I will present the details of my (dis-)proof for the inequality "m < p^k", which was the last stumbling block for my current approach in an attempt to resolve the OPN conjecture.
1 comment:
m < p^k
--> k = 1
--> m < p
--> m < p < Sigma(p) < Sigma(m) < 2m < 2p
--> (1 < (p/m)I(p) < 2) and (1 < (m/p)I(m) < 2)
Let x = (p/m)I(p) and y = (m/p)I(m).
If x = y, then Sigma(p)/m = Sigma(m)/p, which implies that: p*Sigma(p) = m*Sigma(m).
Since gcd(p, m) = 1, we have:
(*) p divides Sigma(m); AND
(**) m divides Sigma(p).
But this means that x = Sigma(p)/m and
y = Sigma(m)/p are both integers, contradicting 1 < x < 2 and 1 < y < 2.
Thus, it is true that either x < y or y < x.
[MORE LATER...]
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