Why Does a 2x2 Matrix [x y; y z] in the PSD Cone Imply x>=0, z>=0, and xz>=y^2?

  • Thread starter Thread starter peterlam
  • Start date Start date
  • Tags Tags
    Cone Positive
peterlam
Messages
16
Reaction score
0
Hi!

If we have a 2x2 matrix [x y;y z] belonging to a positive semidefinite cone. Why is it equivalent to say x>=0, z>=0, and xz>=y^2?

Thanks!
 
Physics news on Phys.org
A matrix is PSD if and only if it's principal minors are nonnegative (see http://en.wikipedia.org/wiki/Positive-definite_matrix#Characterizations").

A 2x2 matrix has three principal minors - the diagonal elements, and the determinant. So x,z >= 0, and xz - y^2 >=0.

I'm sure there is a way to see this without having to use the principal minors characterization though.
 
Last edited by a moderator:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top