35
$\begingroup$

$\DeclareMathOperator\Prob{Prob}$Let $X,Y\in M_n(\mathbb{R})$ be $2$ random matrices. The entries of $X,Y$ are i.i.d. variables. They follow the standard normal law $N(0,1)$.

i) When $n=2,3,4$, one can show that the expectation of $\det(X^2+Y^2)$ is $(n+1)!$.

Computational tests seem to confirm these results for $n\geq5$.

Is the following true ?

Conjecture 1. For every $n\geq 5$, $E(\det(X^2+Y^2))=(n+1)!$.

ii) If $X_0,Y_0$ is the result of an experiment, then let $p_n=\Prob(\det({X_0}^2+{Y_0}^2)<0)$.

Some experiments give the following approximations.

$n=2,p_2\approx 0.1178;n=4,p_4\approx 0.327;n=6,p_6\approx 0.438; n=8,p_8\approx 0.480$; $n=10,p_{10}\approx 0.494;n=20,p_{20}\approx 0.49997;n=40,p_{40}\approx 0.49999$.

Is the following true ?

Conjecture 2. The sequence $(p_n)$ is increasing and tends to $\dfrac{1}{2}$ when $n$ tends to $+\infty$.

If yes and $n$ is great, then $$\Prob(\det({X_0}^2+{Y_0}^2)<0)\approx \Prob(\det({X_0}^2+{Y_0}^2)>0).$$

EDIT. Thanks to everyone who gave a nice answer to conjecture 1.

I don't know if conjecture 1 helps prove conjecture 2.

The problem is that the positive values taken by the determinant are much larger - in absolute value - than the negative ones.

$\bullet$ Note that, since $X$ is "always" invertible

$\Prob(\det(X^2+Y^2)<0)=\Prob(\det(I+X^{-2}Y^2)<0)=\Prob(\det(I+(YX^{-1})(X^{-1}Y))<0)$.

$\bullet$ With the same hypothesis as above concerning the $(X_i)$, it seems that for every $p$ and for $n$ great enough,

$\Prob(\det({X_1}^2+\cdots+{X_p}^2)<0)\approx 1/2$. ​

$\endgroup$
13
  • 6
    $\begingroup$ interesting conjecture; a similar identity, $\mathbb{E}[\det(X+Y)]=(-1)^m (2m)!/m!$ for $n=2m$, was proven by M.K. Chung $\endgroup$ Oct 17, 2023 at 20:48
  • 2
    $\begingroup$ I formulated and verified a conjecture similar to Conjecture 1 for quaternionic matrices. If $A$ is an $m\times n$-quaternionic matrix, then let $\chi_A$ be the corresponding $2m\times 2n$-complex matrix. Then whenever $A$ is a random $n\times n$-quaternionic matrices, we experimentally have $E(\det(\chi_A))=n!$. Here, the entries $v$ in $A$ are Gaussian with independent with mean $0$ and where if $\alpha$ is a unit quaternion, then $\text{Var}(\text{Re}(\alpha v))=1/4$. $\endgroup$ Oct 17, 2023 at 21:22
  • 3
    $\begingroup$ @CarloBeenakker I suppose in your formula $Y$ is supposed to be the transpose of $X$? If $X$ and $Y$ are independent, the determinant vanishes. $\endgroup$ Oct 18, 2023 at 7:24
  • 5
    $\begingroup$ @TimothyBudd --- yes, apologies, $\mathbb{E}[\det(X+X^\top)]=(-1)^m (2m)!/m!$ (this is stated correctly by Chung, it's my mistake) $\endgroup$ Oct 18, 2023 at 8:19
  • 7
    $\begingroup$ More generally I guess $E[\det(X_1^2+X_2^2+\ldots+X_m^2)]=(m+n-1)!/(m-1)!$ for independent matrices $X_i$ of the same type. $\endgroup$
    – Dan Piponi
    Oct 18, 2023 at 17:32

4 Answers 4

21
$\begingroup$

This is easy to do using a graphical calculus for contractions of old-fashioned tensors. See this recent article for an example of application of such techniques and hopefully useful references.

Here is a proof of Conjecture 1. When I have time I will try to make nicer pictures.

Represent a matrix element $M_{ij}$ by a triangular box labelled $M$ and two strands coming out carrying the indices $i$ and $j$, a bit like $i-<M|-j$ (sorry can't complete the two missing corners of the triangle), then see that picture rotated clockwise by 90 degrees.

Then the determinant is given by $$ {\rm det}(M)= \ \ \begin{array}{ccccccc} & & --& -- &-- & & \\ & / & --&-- & --&\backslash & \\ / & / & & & & \backslash & \backslash \\ | & | & & & & | & | \\ | & | & & & & M & M \\ | & | & & & & | & | \\ | & | & & & & = & = \\ | & | & & & & | & | \\ \backslash & \backslash & & & & / & / \\ & \backslash & --& -- &-- & / & \\ & & -- & -- &-- & & \end{array} $$

This picture is for $n=2$. The row of equal signs represents a single box for a normalized antisymmetrizer with $n$ strands going through it. There is also a "Markov trace" or closing the loops without crossings.

Now take $M=X^2+Y^2$ and expand by multilinearity. That means that, in the previous picture, each $$ \begin{array}{c} | \\ M \\ | \end{array} $$ becomes $$ \begin{array}{c} | \\ X \\ | \\ X \\ | \end{array} $$ or $$ \begin{array}{c} | \\ Y \\ | \\ Y \\ | \end{array} $$ Now the placement of the pairs of $X$'s and $Y$'s does not matter, because exchanging positions creates two twists, one on top and one below the antisymmetrizer. Undoing them gives a factor $(-1)^2$, i.e., does nothing. So we get a sum $$ \sum_{n_1+n_2=n}\frac{n!}{n_1! n_2!}\times $$ the expectation of the picture above with the $M$'s (now $n$ of them) are replaced by something like $$ \begin{array}{cccccccc} | & | & \cdots & | & | & | & \cdots & | \\ X & X & \cdots & X & Y & Y & \cdots & Y \\ | & | & \cdots & | & | & | & \cdots & | \\ X & X & \cdots & X & Y & Y & \cdots & Y \\ | & | & \cdots & | & | & | & \cdots & | \end{array} $$ where there are $n_1$ vertical strands of $X$'s followed by $n_2$ strands of $Y$'s.

Now we apply the Isserlis-Wick Theorem in order to perform the Gaussian integrals, as a sum over perfect matchings of $X$'s among themselves and of $Y$'s among themselves. The main ingredient is the graphical identity $$ \mathbb{E}\left[ \begin{array}{c} | \\ X \\ | \end{array} \begin{array}{c} | \\ X \\ | \end{array} \right]= \begin{array}{} | & & | \\ - & -- & - \\ & & \\ - & -- & - \\ | & & | \end{array} $$ namely, with indices, $$ \mathbb{E}[X_{ij}X_{k\ell}]=\delta_{ik}\delta_{j\ell} $$ A better picture than using "$--$"'s and "$|$" would show a cup $\bigcup$ on top and a cap $\bigcap$ on the bottom.

Notice that two $X$'s in the same row cannot contract, because otherwise we would get a symmetric Kronecker delta contracted directly with an antisymmetrizer and this gives zero. So the $i$-th $X$ in the bottom row should contract with the $\sigma_1(i)$-th $X$ in the top row and the $j$-th $Y$ in the bottom row should contract with the $\sigma_2(j)$-th $Y$ in the top row. Summing over Wick contractions thus reduces to summing over two permutations $\sigma_1\in S_{n_1}$ and $\sigma_2\in S_{n_2}$.

Now the hard part, and sorry the pictures get a bit tricky (try to draw a cycle of length four), the contribution of the permutations can be computed as follows. After straightening or pulling on the strands, one gets the compositional squares of the two permutations. In the Schur-Weyl picture we would have something like $$ \omega= (\sigma_1\circ \sigma_1) \otimes (\sigma_2\circ\sigma_2) $$ as an endomorphism of $\wedge^n(\mathbb{R}^n)$.

Another way to see this is as follows. Consider the $i$-th "$|$" below a bottom row $X$. It will follow a cap going to the strand "$|$" right below the $\sigma_1(i)$-th top row $X$. You have to go down that strand and then follow the cup that takes you to the "$|$" sitting on top of the $\sigma_1(\sigma_1(i))$-th top row $X$. At the end of the day after pulling on the strands, eliminating curlicues $X|$ and zigzags $/|/$, you have that the $i$-th strand at the very bottom connects to the $\sigma_1(\sigma_1(i))$-th strand at the very top. That was for the $X$ compartment, but the same is done independently in the $Y$ compartment.

Then the evaluation of the the contribution of $\omega$ is the sign of the permutation $\omega$, i.e., $1$ times the trace of the identity on $\wedge^n(\mathbb{R}^n)$ which is also $1$.

In sum, and for the general case of $m$ matrices, we have $$ \mathbb{E}[{\rm det}(X_1^2+\cdots+X_m^2)]= \sum_{n_1+\cdots+n_m=n} \binom{n}{n_1,\ldots,n_m} n_1!\cdots n_m!\ \times 1 $$ $$ =n!\sum_{n_1+\cdots+n_m=n} 1=n!\times\binom{n+m-1}{m-1}=\frac{(n+m-1)!}{(m-1)!}\ . $$ This proves the more general conjecture in Dan Piponi's comment.

$\endgroup$
7
  • 8
    $\begingroup$ Can someone explain to me what on Earth is going on in this answer? Any references appreciated. $\endgroup$ Oct 18, 2023 at 22:10
  • 5
    $\begingroup$ See, e.g., the book by Cvitanovic birdtracks.eu/version9.0/GroupTheory.pdf If you have never seen this, you have to practice a bit. $\endgroup$ Oct 18, 2023 at 23:21
  • 1
    $\begingroup$ Can this method be used to compute $\Bbb E\det\sum X_i^k$ for integer $k>2$? $\endgroup$ Oct 19, 2023 at 10:40
  • 3
    $\begingroup$ There is a nice generating function for $\mathbb{E}\det X^4$ so maybe this can be extended to $\mathbb{E}\det \sum X_i^4$. For $\mathbb{E}\det X^6$ see arXiv:2206.11356. It looks like $\mathbb{E} \det X^8$ will be very complicated. $\endgroup$ Oct 19, 2023 at 13:21
  • 2
    $\begingroup$ @TheSimpliFire: The beginning and the end of the apply. The computation of the contribution of $\omega$ is also the sign of that permutation. But one would also get isolated/disconnected loops contributing a factor of $n$ each. The part which would likely be too complicated is counting the analogues of the permutations $\sigma_i$ which would become much more complicated combinatorial objects. Follwing up on Richard's comment, one could start by looking up the article by Beck arxiv.org/abs/2207.09311 $\endgroup$ Oct 19, 2023 at 15:45
19
$\begingroup$

This is the translation of the @AbdelmalekAbdesselam answer to the standard algebraic notation, for those who, like me, feel not so comfortable with birdtracks notation. The main idea is to expand $\det(X^2 + Y^2)$ in the sum of monomials on variables $X_{ij},Y_{ij}$ and then apply Isserlis' formula. Namely, one has $$\mathbb E[\det(X^2+Y^2)] = \sum_{\sigma \in \Sigma_n} (-1)^{\sigma} \mathbb E[(a_{1,\sigma 1} + b_{1,\sigma 1}) \cdots (a_{n,\sigma n} + b_{n,\sigma n})] =$$ where $A = X^2, B=Y^2$. Now, after expanding and regrouping $a$-terms with b-terms and using various symmetries of the problem $$=\sum_{\sigma \in \Sigma_n}\sum_{\substack{n_1 + n_2 = n \\\{i_1,\ldots,i_{n_1}\} \sqcup \{j_1,\ldots,j_{n_2}\} = \{1,\ldots,n\} \\ i_1 < \cdots < i_{n_1}\\j_1 < \cdots < j_{n_2}}} (-1)^{\sigma} \mathbb E[a_{i_1,\sigma i_1} \cdots a_{i_{n_1}, \sigma i_{n_1}}]\mathbb E[b_{j_1,\sigma j_1} \cdots b_{j_{n_2}, \sigma j_{n_2}}] =$$ $$=\sum_{n_1+n_2=n}\frac{n!}{n_1! n_2!} \sum_{\sigma_1 \in \Sigma_{n_1}, \sigma_2 \in \Sigma_{n_2}} (-1)^{\sigma_1} (-1)^{\sigma_2} \mathbb E[a_{1,\sigma_1 1} \cdots a_{n_1, \sigma n_1}]\mathbb E[b_{1,\sigma_2 1} \cdots b_{n_2, \sigma_2 n_2}] =$$ $$=\sum_{n_1 + n_2 = n} \frac{n!}{n_1! n_2!} \mathbb E[\det(Z_{n_1}^2)] \mathbb E[\det(Z_{n_2}^2)]$$

where $Z_r$ is an $r \times r$ matrix filled with iid normal distributions. So it is enough to prove

$$\mathbb E[\det(Z_{r})^2]=r!$$ Via Isserlis' formula only quadratic monomials of $\det(Z_{r})^2$ will contribute to the answer, and there are exactly $r!$ of them.

$\endgroup$
6
$\begingroup$

Not (yet?) a complete answer


For conjecture 1, it is helpful to represent the determinant of an $n\times n$ matrix $M$ as an integral over anticommuting (Grassmann) variables $\theta=(\theta_1,\theta_2,\ldots\theta_n)$, and their conjugates $\bar{\theta}=(\bar{\theta}_1,\bar{\theta}_2,\ldots\bar{\theta}_n)$,

$$\det M=\int d\theta\int d\bar{\theta}\,e^{\bar{\theta}\cdot M\cdot\theta}=\int d\theta\int d\bar{\theta}\,\prod_{i=1}^n\left(1+\bar{\theta}_i\sum_{j=1}^n M_{ij}\theta_j\right),\tag{1}$$ as explained, for example, in these lecture notes.

Apply this to $M=X^2+Y^2$, $$\det(X^2+Y^2)=\int d\theta\int d\bar{\theta}\,e^{\bar{\theta}\cdot X^2\cdot\theta}e^{\bar{\theta}\cdot Y^2\cdot\theta}$$ $$\qquad=\int d\theta\int d\bar{\theta}\,\prod_{i,i'=1}^n\left(1+\bar{\theta}_i\sum_{j,k=1}^n X_{ik}X_{kj}\theta_j\right)\left(1+\bar{\theta}_{i'}\sum_{j',k'=1}^n Y_{i'k'}Y_{k'j'}\theta_{j'}\right).\tag{2}$$

So I need to evaluate a Gaussian average of the form $$Z_n(\bar{\theta},\theta)=\mathbb{E}\biggl[\prod_{i=1}^n\biggl(1+\sum_{j,k=1}^n X_{ik}X_{kj}\bar{\theta}_i\theta_j\biggr)\biggr]\tag{3}$$ and then perform the remaining integral of $Z_n^2$ over Grassmann variables, $$\mathbb{E}[\det(X^2+Y^2)]=\int d\theta\int d\bar{\theta}\,Z_{n}^2(\bar{\theta},\theta),\tag{4}$$ with the help of the identities $$\int d\theta_id\bar{\theta}_i=0,\;\;\int d\theta_id\bar{\theta}_i\,\bar\theta_i=0,\;\int d\theta_id\bar{\theta}_i\,\bar\theta_i\theta_i=1.\tag{5}$$

Similarly, the multi-matrix generalization of Dan Piponi corresponds to $$\mathbb{E}\left[\det\left(\sum_{i=1}^m X_i^2\right)\right]=\int d\theta\int d\bar{\theta}\,Z_{n}^m(\bar{\theta},\theta).\tag{6}$$


As a quick check that this is leading somewhere, for $n=2$ one has $Z_2=1+\bar{\theta}_{1}\theta_1+\bar{\theta}_{2}\theta_2+2\bar{\theta}_{1}\theta_1\bar{\theta}_{2}\theta_2$, giving $$\mathbb{E}[\det(X^2+Y^2)]=\int d\theta\int d\bar{\theta}\,\biggl(1+\bar{\theta}_{1}\theta_1+\bar{\theta}_{2}\theta_2+2\bar{\theta}_{1}\theta_1\bar{\theta}_{2}\theta_2\biggr)^2$$ $$\qquad=\int d\theta\int d\bar{\theta}\,\biggl(1+2\bar{\theta}_{1}\theta_1+2\bar{\theta}_{2}\theta_2+6\bar{\theta}_{1}\theta_1\bar{\theta}_{2}\theta_2\biggr)=6,$$ which is the correct answer.

The $m$-matrix generalization of Dan Piponi evaluates for $n=2$ and arbitrary $m$ to $$\mathbb{E}\left[\det\left(\sum_{i=1}^m X_m^2\right)\right]=\int d\theta\int d\bar{\theta}\,\biggl(1+\bar{\theta}_{1}\theta_1+\bar{\theta}_{2}\theta_2+2\bar{\theta}_{1}\theta_1\bar{\theta}_{2}\theta_2\biggr)^m=m(m+1), $$ because only the term $\bar{\theta}_{1}\theta_1\bar{\theta}_{2}\theta_2$ gives a nonzero value upon integration over the Grassmann variables, and this term appears with multiplicity $2m+m(m-1)=m(m+1)$.

$\endgroup$
5
$\begingroup$

I'd like to expand on my comment above to give a more pedestrian proof which also connects to some other standard facts in combinatorics.

Start with $$ E[\det(X^2)] =\sum_{\sigma\in S_n}(-1)^{\mathop{sign(\sigma)}}\prod_{i=1}^n(X^2)_{i,\sigma(i)} =\sum_{\sigma\in S_n}(-1)^{\mathop{sign(\sigma)}}\prod_{i=1}^n\sum_{j=1}^n X_{i,j}X_{j,\sigma(i)} $$ The only terms on the right that survive are those that are products of matching pairs of variables. Each surviving term on the right is a product of variables that I'll write in a grid rather than in a line. Eg. $$ \begin{array}{cc} X_{1,j_1} & X_{j_1,\sigma(1)} \\ X_{2,j_2} & X_{j_2,\sigma(2)} \\ \vdots \\ X_{n,j_n} & X_{j_n,\sigma(n)} \\ \end{array} $$ For every variable to appear precisely twice, the right column must in fact be a permutation of the left column. So the function $i\mapsto j_i$ is a permutation that I'll write as $j_n=j(n)$ and we have $\sigma=j^2$. So in fact we have $$ E[\det(X^2)]=\sum_{j\in S_n}(-1)^{sign(j^2)}1=n! $$ Now consider $E[\det(X^2+Y^2)]$. Like before we have a sum of terms where each term is a product of $2n$ variables that can be arranged in a grid. For examples some terms are like $$ \begin{array}{cc} X_{1,j_1} & X_{j_1,\sigma(1)} \\ Y_{2,j_2} & Y_{j_2,\sigma(2)} \\ \vdots \\ X_{n,j_n} & X_{j_n,\sigma(n)} \\ \end{array} $$ with some rows containing just $X$'s and some containing just $Y$'s. Now suppose $X_{i,j}$ exists in some row in the left column. Then a variable of the form $X_{j,k}$ must appear in same row but in the right column. The variable $X_{j,k}$ must appear again somewhere on the left. And that means there's a variable of the form $X_{k,l}$ on the right and so on. So cycles must be made up of all $X$'s or all $Y$'s and we can assign $X$ and $Y$ however we like. Similarly for a sum of $m$ squares in $\det(X_1^2+\ldots+X_m^2)$. So now we have $$ E[\det(X_1^2+\ldots+X_m^2)]=\sum_{j\in S_n}m^\mbox{#cycles in $j$}.$$

It's a standard fact that the number of permutations with a given number of cycles is counted by the (unsigned) Stirling numbers of the first kind and that the polynomial you get using these as coefficients is the rising factorial.

So $$ E[\det(X_1^2+\ldots+X_m^2)]=m^{\bar{n}}=\frac{(m+n-1)!}{(m-1)!} $$

The fact that we're looking at a product of terms $X_{i,j}X_{j,\sigma(i)}$ coming from $X^2$ leads to a big constraint so the right column is a permutation of the left. When looking at higher powers of $X$ we get much looser constraints so this approach likely won't work without a lot of extra work.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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