0
$\begingroup$

Let $A \in \mathbb{R}^{n \times n}$ be symmetric and invertible. $K=\{ x \in \mathbb{R}^n : \|Ax\|_2 \le 1 \}$. Now I have to calculate:

$$\int_K \langle A^2x, x \rangle \mathrm d x $$

It exists a orthogonal matrix $S \in \mathbb{R}^{n \times n}$ with $$SAS^T=D \text{ and } SA^2S^T = D^2 \ (D=\text{diag}(\lambda_1,\ldots,\lambda_n))$$

With $\varphi(x)=Sx$ and $B=\{ x \in \mathbb{R}^n : \|Dx\|_2\le 1 \}$ we get $\varphi(B)=K$. Because $|\det(\varphi')|=1$ we get

$$ \int_{\varphi(B)} \langle A^2x, x \rangle \mathrm dx = \int_{B} \langle D^2x,x \rangle \mathrm d x = \int_B \lambda_1^2 x_1^2+\ldots +\lambda_n^2 x_n^2 \mathrm d x $$

At this point I stuck. Is there any further step to simplify this integral?

$\endgroup$
  • $\begingroup$ How would you compute volume of the $n$-dimensional ball $\{x\in \Bbb R^n: \|x\|_2\le 1\}$? $\endgroup$ – Quang Hoang Aug 18 '14 at 16:15
  • $\begingroup$ $n$-dimensional polar coordinates. Seems very messy to do so. $\endgroup$ – DerJFK Aug 18 '14 at 16:23
  • $\begingroup$ This is the exact integral, but you have $r^{n+1}dr$ instead of $r^{n-1}dr$. I don't think there's a better way. $\endgroup$ – Quang Hoang Aug 18 '14 at 16:39
  • $\begingroup$ Why you guys ask the same question? here $\endgroup$ – Troy Woo Aug 18 '14 at 16:40
  • $\begingroup$ Didn't see the other post. $\endgroup$ – DerJFK Aug 18 '14 at 16:43
1
$\begingroup$

Let $\phi(x) = A^{-1} x$, and $f(x) = \|Ax\|^2$.

Note that $K=\{x | \|Ax\| \le 1 \} = \{ A^{-1}y | \|y \| \le 1 \} = \phi(B(0,1))$.

$\int_K \|Ax\|^2 dx = \int_{\phi(B(0,1))} f = \int_{B(0,1)} f \circ \phi |J_\phi| = {1 \over \det A} \int_{B(0,1)} \|x\|^2 dx$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.