Motivation. If $A \in \mathbb{C}^{n \times n}$ is self-adjoint (or, more generally, normal), then we all know that $$ A = \sum_{k=1}^n \lambda_k \, h_k \otimes h_k, $$ where $\lambda_1,\dots,\lambda_n$ are the eigenvalues of $A$ (counting multiplicities), $(h_1,\dots,h_n)$ is a corresponding orthonormal basis of eigenvectors, and $h_k \otimes h_k$ denotes the matrix given by $(h_k \otimes h_k)x = \langle h_k, x\rangle h_k$ for each $x \in \mathbb{C}^n$ (here I used the "physical" convention that the inner product is linear in the second component).

A representation result. The following result for general (i.e. also non-normal) matrices is - very loosely - reminiscent of the above quoted spectral theorem:

Let $S$ denote the (Euclidean) unit sphere in $\mathbb{C}^n$ and let $\lambda$ denote the surface measure on $S$ (more precisely, we identify $\mathbb{C}^n$ with $\mathbb{R}^{2n}$, consider the surface measure on the unit sphere there and pull it back to $S$). Now, normalize $\lambda$ such that $\lambda(S) = n$.

Theorem. For every matrix $A \in \mathbb{C}^{n\times n}$ we have $$ A = (n+1) \int_{S} \langle h, Ah \rangle \; h \otimes h \; d \lambda(h) - \operatorname{tr}(A) \, I; $$ here, $\operatorname{tr}(A)$ denotes the trace of $A$ and $I \in \mathbb{C}^{n\times n}$ denotes the identity matrix.

One can prove the above theorem by using the answers to this MathOverflow question.

The question (a reference request). I have no idea whether the above representation theorem is of any use - but given its very symmetric and rather simple nature, it is natural to suspect that the theorem should already be somewhere in the literature.

So my question is: Do you know any reference where the above representation theorem is stated and proved?

  • $\begingroup$ @FrancoisZiegler: Well, it's precisely the computation of those two scalars that took me a page or two. But admittedly, "lengthy" might be a bit of an exagaration. I think I'll better remove this word from the post. $\endgroup$ Commented Feb 9, 2020 at 23:31
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    $\begingroup$ If I'm not mistaken then it's also possible to use the coarea formula to rewrite the integral over $S^{2n-1}$ into a double integral over the unit disk and $S^{2n-3}$. An induction over $n$ then should do the trick. Alternatively: Pick a basis and do everything in coordinates. But this might be "lengthy" again. $\endgroup$ Commented Feb 9, 2020 at 23:33
  • $\begingroup$ @JohannesHahn: Thank you for your comment! I agree that one can probably show the formula by concretely computing some integrals. The major motivation of my question though was not to look for an alternative proof (probably the original wording of the question was somewhat misleading, and suggested that I was looking for a simpler or less "lengthy" proof). The main reason for my reference request was that I was wondering in which contexts this formula might occur in the literature. $\endgroup$ Commented Feb 10, 2020 at 16:42
  • $\begingroup$ @FrancoisZiegler: Thank you very much for the references! (I particularly like the first one.) If you post the reference as an answer, I'll certainly accept it. $\endgroup$ Commented Feb 10, 2020 at 16:43

3 Answers 3


My comments converted to an answer:

1st comment. I know no reference but the proof need not be [as per OP] lengthy — one just computes the two scalars by which the map $\mathscr I:$ $$ \textstyle A\mapsto\mathscr I(A)=\int_S h\langle h,Ah\rangle\langle h,\cdot\rangle d\lambda(h) \tag1 $$ acts on the irreducible components of $\mathfrak{gl}(n,\mathbf C)=\mathfrak{sl}(n,\mathbf C)\oplus\mathbf C I$ (Schur’s lemma).

(In detail: write $r$ and $s$ for the scalars in question. Taking the trace of $sI=\mathscr I(I)$ in (1) gives $s=1$. It follows that we have $ \mathscr I(A)= r\left(A-\tfrac{\operatorname{tr}A}{n}I\right)+\tfrac{\operatorname{tr}A}{n}I $ and hence $$ \operatorname{tr}(\mathscr I(A)B)=r\operatorname{tr}(AB)+\tfrac{1-r}n\operatorname{tr}(A)\operatorname{tr}(B)\rlap{\qquad\quad\forall A, B.} \tag2 $$ Writing this out for $(A,B)=(E_{12},E_{21})$, resp. $(E_{11},E_{22})$ where $E_{ij}=e_i\langle e_j,\cdot\rangle$, one obtains that $\int_S|\langle e_1,h\rangle|^2|\langle e_2,h\rangle|^2d\lambda(h)$ equals both $r$ and $\frac{1-r}n$. Therefore $r=\frac1{n+1}$, and with that (2) becomes (3) below. QED)

2nd comment. For a reference: your desired formula is equivalent to $$ \mathrm{tr}(AB)+\mathrm{tr}(A)\mathrm{tr}(B)=(n+1)\int_S\langle h,Ah\rangle\langle h,Bh\rangle\,d\lambda(h) \tag3 $$ which is e.g. (3.8) of Gibbons (1992), taking differences of notation into account. (The integrand descends to $P^{n-1}(\mathbf C)$ where he uses the measure of total volume $\pi^{n-1}/(n-1)!$) I think it should also follow from Archimedes-Duistermaat-Heckman (1982, Prop. 3.2).

More context: Generally I think you’ll find many similar formulas in the (somewhat repetitive) literature on “coherent states” or “quantum mechanics as classical mechanics on $P\mathscr H$”. Also compare Schur’s proof of his orthogonality relations (1924, p.199 or Bröcker-tom Dieck 4.5i), specialized to the adjoint representation.


This is an easy consequence of the $k=2$ case of the complex version of the Isserlis-Wick theorem for moments of Gaussian measures, i.e., the identity $$ \int_{\mathbb{C}^n} z_{i_1}\cdots z_{i_k}\ {\bar{z}}_{j_1}\cdots {\bar{z}}_{j_k} \ e^{-|z|^2} \prod_{a=1}^{n}\frac{d(\Re z_a) d(\Im z_a)}{\pi}\ =\ \sum_{\sigma\in\mathfrak{S}_k} \delta_{i_1 j_{\sigma(1)}}\cdots \delta_{i_k j_{\sigma(k)}}\ . $$

Going to spherical coordinates produces the integral $\int_{S}\cdots d\lambda(h)$, while the sum over the permutation $\sigma$ gives the other two terms of the wanted identity.

Indeed, it is good to see this really as an integral over $\mathbb{C}\mathbb{P}^{n-1}$ for the Fubini-Study metric, but appealing to Duistermaat-Heckman is not necessary (as known to François).

  • $\begingroup$ That’s essentially Gibbons’ proof. Mine (“3rd comment”) uses neither Duistermaat-Heckman (mentioned for context) nor really any integral formula. $\endgroup$ Commented Feb 10, 2020 at 22:42
  • $\begingroup$ Well, yes and no. For $V=\mathbb{C}^n$, finding that the space of $U(n)$ invariants in $V\otimes V\otimes V^{\vee}\otimes V^{\vee}$ is spanned by two elements can be done by averaging over the group. See my two answers to mathoverflow.net/questions/255492/… So I would say an integral formula is hiding behind the representation theoretic proof. Once that is known, finding the scalars by contracting with suitable "test functions" is not hard. $\endgroup$ Commented Feb 10, 2020 at 23:02
  • $\begingroup$ Don’t get me wrong, I like this too (and, someone should write “Elementary mathematics from an Atiyah-Singer standpoint”). But I felt like I managed with nothing more than $d\lambda$’s invariance + Schur’s lemma, so now I’m curious: at which precise step do you see me smuggling in any computed integral at all? $\endgroup$ Commented Feb 11, 2020 at 4:22
  • $\begingroup$ A proof of course must start somewhere. Yours starts from some basic results in representation theory, whereas mine begins from a more rudimentary starting point, namely, multivariate calculus. The way I see it, the hidden use of integrals is in the used results from representation theory, not in the follow up explicitly written in your answer. Look at I how I prove Schur-Weyl duality in the link in the comment above and you will see what I mean. $\endgroup$ Commented Feb 11, 2020 at 14:44
  • $\begingroup$ Food for thought, thanks. [Later:] I think I see now. A computation is hidden, not in the representation theory, but in the OP’s phrase “Now, normalize 𝜆 such that 𝜆(𝑆) = 𝑛.” (This cannot be done without computing 𝜆(𝑆).) What we can prove with absolutely no integral evaluated is: for any 𝑈(𝑛)-invariant 𝜇 on 𝑆,$$A+\mathrm{tr}(A)I=\frac{n(n+1)}{\mu(S)}\int_S h\langle h,Ah\rangle\langle h,\cdot\rangle\,d\mu(h).$$ $\endgroup$ Commented Feb 17, 2020 at 2:45

Here is a down-to-earth proof. Remark that for complex matrices, proving $A=B$ is equivalent to proving $\langle x,Ax\rangle=\langle x,Bx\rangle$ ; the superiority of the complex numbers over the real ones !

We must prove that $$\langle x,Ax\rangle=(n+1)\int_S|\langle h,x\rangle|^2\langle h,Ah\rangle d\lambda(h)-({\rm Tr}\,A)|x|^2.$$ By rotational invariance, it is enough to prove this for $x=\vec e_1$, that is $$a_{11}=(n+1)\int_S|h_1|^2\langle h,Ah\rangle d\lambda(h)-{\rm Tr}\,A.$$ This amounts to verifying the following identities, all of which being classical: $$\int_S|h_1|^4d\lambda(h)=\frac2{n+1},\quad\int_S|h_1|^2|h_2|^2d\lambda(h)=\frac1{n+1}$$ and $$\int_S|h_1|^2h_j\bar h_kd\lambda(h)=0,\quad j\ne k.$$

  • $\begingroup$ Thank you for your answer! If I could, I would give you a second upvote for "the superiority of the complex numbers over the real ones!" $\endgroup$ Commented Feb 19, 2020 at 12:50
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    $\begingroup$ @JochenGlueck. It deserves to be more known. This property is equivalent to saying that the numerical radius is a norm over ${\bf M}_n({\mathbb C})$. Even if a matrix has real entries, it is crucial to compute its numerical radius from complex vectors, not only real ones ! $\endgroup$ Commented Feb 19, 2020 at 13:01

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