Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For arbitrary $n\times n$ matrices M, I am trying to solve the integral

$$\int_{\|v\| = 1} v^T M v.$$

Solving this integral in a few low dimensions (by passing to spherical coordinates) suggests the answer in general to be $$\frac{A\,\mathrm{tr}(M)}{n}$$ where $A$ is the surface area of the $(n-1)$-dimensional sphere. Is there a nice, coordinate-free approach to proving this formula?

share|improve this question
I would try proving this for diagonal matrices and arguing that the integral doesn't change when you replace $M$ by $PMP^{-1}$. –  Grumpy Parsnip Sep 6 '11 at 17:57
@Jim: It says "arbitrary", not just diagonalizable. –  joriki Sep 6 '11 at 17:57
@Joriki: that's true. At least that would prove a special case. :) –  Grumpy Parsnip Sep 6 '11 at 18:00
@Jim: Thanks! That definitely seems like the right track (though I think the integral only doesn't change when $P$ is orthogonal, i.e. when $M$ is symmetric) –  user7530 Sep 6 '11 at 18:05
Of course Jim's comment also leads to a slightly more heavy-handed answer by replacing the integral with $\frac{1}{n}\int_{SO(n)} \mathrm{tr}(OMO^{-1})$. –  Willie Wong Sep 6 '11 at 18:09
add comment

2 Answers

up vote 9 down vote accepted

The integral is linear in $M$, so we only have to calculate it for canonical matrices $A_{kl}$ spanning the space of matrices, with $(A_{kl})_{ij}=\delta_{ik}\delta_{jl}$. The integral vanishes by symmetry for $k\neq l$, since for every point on the sphere with coordinates $x_k$ and $x_l$ there's one with $x_k$ and $-x_l$.

So we only have to calculate the integral for $k=l$. By symmetry, this is independent of $k$, so it's just $1/n$ of the integral for $M$ the identity. But that's just the integral over $1$, which is the surface area $A$ of the sphere.

Then by linearity the integral for arbitrary $M$ is the sum of the diagonal elements, i.e. the trace, times the coefficient $A/n$.

share|improve this answer
@Willie Indeed there is. Corrected, thanks. –  user7530 Sep 6 '11 at 18:09
Very nice. $\,\,\,$ –  Grumpy Parsnip Sep 6 '11 at 18:09
add comment

As a function of $M$ your integral is linear, and is invariant under conjugation by orthogonal transformations ($C_R: M \mapsto R^{T} M R$). Now the average of $C_R(M)$ over all orthogonal transformations $R$ (using Haar measure) is $\text{Tr}(M) I/n$ (it must be invariant under all $C_R$ so it is a multiple of $I$, and the trace is preserved). So the integral is the same as it would be for $\text{Tr}(M)I/n$, which is $\text{Tr}(M)/n$ times the area of the sphere.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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