Comparing Matrices: Understanding Notation and Inequalities

dlh81
Messages
1
Reaction score
0
I am having an issue with a problem, but mostly because I am confused by the notation. This is the question:

If Q is an n x n symmetric matrix and e1, e2 (e is epsilon) are such that
0 < e1*I </= Q </= e2*I

show that

1/e2 * I </= Q^-1 </= 1/e1 *I

( </= is less or equal to)



My question is how can a matrix be compared with another matrix in a quantitative manner (less than, less or equal to, greater than, etc.) I am familiar with norms or determinants being compared that way since they are a scalar, but how would n x n matricies be compared this way? Any suggestions?

Also I am assuming that the epsilons are scalar multipliers. The book I am using does not do a good job of clarifying this notation, but that is all I can imagine it would be.
 
Physics news on Phys.org
A<=B means that (B-A) is positive semi-definite. Of course this makes sense only when A, B are hermitian matrixes.
 
On a side note, <= is widely understood to mean "less than or equal to" and >= is understood to mean "greater than or equal to."
 
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