1
$\begingroup$

i have troubles to understand this computation: Let $A=(a_{ij})_{i,j=1}^n$ be a real matrix. It holds $$ \int_{B_1(0)}Ax\cdot x dx=\int_{B_1(0)} \sum_{i,j=1}^na_{ij}x_ix_jdx=\int_{B_1(0)} \sum_{i=1}^na_{ii}x_i^2dx=\dfrac{1}{n}trace(A)\int_{B_1(0)}\vert x\vert^2dx=\dfrac{\omega_n}{n(n+2)}trace(A) $$ where $\omega_n$ is the area of the unit sphere in $\mathbb{R}^n$. First step is obvious. In the second step they use $\int_{B_1(0)} a_{ij}x_ix_jdx=0$ for $i\neq j$. It easy to show this for $n=1,2,3$. But for general $n$ i need help. Third step is also clear. They use $\int_{B_1(0)}x_1^2dx=\ldots=\int_{B_1(0)}x_n^2dx$. Then, we have $$ \int_{B_1(0)}x_i^2dx=\dfrac{1}{n}\int_{B_1(0)}\vert x\vert^2dx $$ for every $i=1,...,n$. Final step is also clear. They use $$ \int_{B_1(0)}\vert x\vert^2dx=\int_0^1 s^2\int_{\partial B_s(0)}dH^{n-1}ds=\omega_n\int_0^1 s^2\cdot s^{n-1}ds=\dfrac{\omega_n}{n+2}. $$ The only thing what i don't understand is $\int_{B_1(0)} a_{ij}x_ix_jdx=0$ for $i\neq j$!

Can anybody explain this?

Thank you.

Best regards

$\endgroup$

1 Answer 1

0
$\begingroup$

By symmetry: \begin{align*} \iint_{B_1(0)} a_{ij} x_i x_j \,\mathrm{d} x_i \,\mathrm{d}x_j &= \iint_{B_+} a_{ij} x_i x_j \,\mathrm{d} x_i \,\mathrm{d}x_j + \iint_{B_-} a_{ij} x_i x_j \,\mathrm{d} x_i \,\mathrm{d}x_j \\ &= \iint_{B_+} a_{ij} x_i x_j \,\mathrm{d} x_i \,\mathrm{d}x_j - \iint_{B_+} a_{ij} x_i x_j \,\mathrm{d} x_i \,\mathrm{d}x_j = 0 \end{align*}

where $B_+ := \{x \in B_1(0) : x_i > 0\}$ and $B_- := \{ x \in B_1(0) : x_i \leq 0 \}$.

$\endgroup$
1
  • $\begingroup$ Wow, thank you very much. $\endgroup$
    – Serdar
    Commented Sep 26, 2017 at 9:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .