This question is not about the proof but about a reliable source where one can find the following formula for the normalised trace $\mbox{tr}$ of a complex $(n\times n)$-matrix:

$$\mbox{tr}(A) = \int\limits_{\|y\|=1} \langle Ax, x\rangle\, \mbox{d}m(x)$$

where $m$ is the Lebesgue (area) measure normalised to the Euclidean sphere in $\mathbb{C}^n$. Thank you.

  • $\begingroup$ Just for clarification, do you mean $dm(x) = \nu(x) d\lambda(x) = \frac{x}{\Vert x\Vert} d\lambda(x)$? $\endgroup$
    – AlexR
    Dec 4, 2013 at 13:34
  • $\begingroup$ Yes, in other words $m$ normalises the Lebesgue measure so that the right hand side gives you 1 for $A=I$, the identity matrix. $\endgroup$ Dec 4, 2013 at 13:36
  • $\begingroup$ @TomekKania: I am rather doubtful that MO is a better place for this question. Besides, the question has only been posted for 22 hours. We'll at least wait a bit before we migrate the question. (In the mean time we'll keep the flag open as a reminder.) $\endgroup$ Dec 5, 2013 at 12:01
  • $\begingroup$ Also, a similar question was asked recently on MO and put on-hold, but an existing answer gives at least for a reference "Riemannian geometry" by Gallot, Hulin, and Lafontaine. The linearity of the trace implies you can split to real and imaginary parts, and the $\mathbb{C}^n$ case should follow from the $\mathbb{R}^n$ case by symmetry arguments. $\endgroup$ Dec 5, 2013 at 12:08
  • $\begingroup$ @TomekKania: I have not had a chance to look at the book of Gallot, Hulin, and Lafontain. If you found that book, or the answer on MO, to be useful, I encourage you to post an answer yourself to this question with the reference you found! $\endgroup$ Dec 6, 2013 at 12:17

2 Answers 2


Note: the following is originally due to Bennett Chow on MathOverflow. Reposting here as CW in view of this comment.

Let $S^{n-1}$ be the unit sphere in some tangent space with inner product $g$. Let $\{e_i\}_{i=1}^n$ be an orthonormal frame and let $V_i=\langle V,e_i\rangle$. For $i\neq j$, $\int_{S^{n-1}}V_{i}V_{j}\operatorname{dvol}\left( V\right) =0$ since the integrand is odd with respect to reflection about the coordinate hyperplane $\left\{ V_{i}=0\right\} $. Taking $i=j$, we get $\int_{S^{n-1}}V_{i}^{2}\operatorname{dvol}\left( V\right) =\frac{\omega_{n}}{n}$ since this expression is independent of $i$ and since $\sum_{i=1}^{n}\int_{S^{n-1}}V_{i}^{2}\operatorname{dvol}\left( V\right) =\int_{S^{n-1} }\operatorname{dvol}\left( V\right) =\omega_{n}$, using $|V|^2=1$. We conclude with $\alpha_{ij}=\alpha (e_i,e_j)$ that $$ \int_{S^{n-1}}\alpha\left( V,V\right) \operatorname{dvol}\left( V\right) =\sum _{i,j=1}^{n}\int_{S^{n-1}}\alpha_{ij}V_{i}V_{j}\operatorname{dvol}\left( V\right) =\sum_{i=1}^{n}\int_{S^{n-1}}\alpha_{ii}V_{i}^{2}\operatorname{dvol}\left( V\right) =\frac{\omega_{n}}{n}\operatorname{Trace}{}_{g}(\alpha). $$


Let me add some references then:

A short proof of the fact that the matrix trace is the expectation of the numerical values, The American Mathematical Monthly 122 (2015), no. 8, 782–783, arXiv version.

and a generalisation to certain other unit spheres:

T. Kania and K. E. Morrison, The trace as an average over the unit sphere of a normed space with a 1-symmetric basis, arXiv version.


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