Showing something satisfies Inner Product - Involves Orthogonal Matrices

Circular_Block
Messages
2
Reaction score
0

Homework Statement


[/B]
Let Z be any 3×3 orthogonal matrix and let A = Z-1DZ where D is a diagonal matrix with positive integers along its diagonal.
Show that the product <x, y> A = x · Ay is an inner product for R3.

Homework Equations


None

The Attempt at a Solution



I've shown that x · Dy is an inner product. I know that Z-1 is equal to ZT. I believe that will lead me somewhere. I'm just having trouble showing the property <x, x> ≥ 0. I also know that (ATx)⋅x = x ⋅ Ax.

Just missing one step. I don't know what it is.
 
Last edited:
Physics news on Phys.org
Hint: ## (Zx)^T = x^T Z^T ##
 
Doing that you'll end up with (ATx)⋅x right?
 
Last edited:
Circular_Block said:
Doing that you'll end up with (ATx)⋅x right?

x.y=x^Ty. That turns a dot product into a matrix product. Add that to the list of clues.
 
Don't neglect that D is all positive. So ##Z^T D Z## should also be non-negative, right?
 
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