Return to Answer

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 $Z_2=1+c_{11}+c_{22}+2c_{11}c_{22}$, 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 $2m+m(m-1)=m(m+1)$.

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 $2m+m(m-1)=m(m+1)$.

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 $Z_2=1+c_{11}+c_{22}+2c_{11}c_{22}$, 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.

Similarly, 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 $2m+m(m-1)=m(m+1)$.

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 $Z_2=1+c_{11}+c_{22}+2c_{11}c_{22}$, 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 $2m+m(m-1)=m(m+1)$.

This does not seem to be an easy road to an answer, I will leave it for the record (or delete it if so requested)

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 $M=X^2+Y^2$, $$\det(X^2+Y^2)=\int d\theta\int d\bar{\theta}\,e^{\bar{\theta}\cdot X^2\cdot\theta}e^{\bar{\theta}\cdot Y^2\cdot\theta}$$ $$\qquad=\int d\theta\int d\bar{\theta}\,\prod_{i,i'=1}^n\left(1+\bar{\theta}_i\sum_{j,k=1}^n X_{ik}X_{kj}\theta_j\right)\left(1+\bar{\theta}_{i'}\sum_{j',k'=1}^n Y_{i'k'}Y_{k'j'}\theta_{j'}\right).\tag{2}$$ We now take the expectation value over the independent normally distributed matrix elements of $X$ and $Y$, $$\mathbb{E}[\det(X^2+Y^2)]=\int d\theta\int d\bar{\theta}\,\left(\mathbb{E}\biggl[\prod_{i=1}^n\biggl(1+\bar{\theta}_i\sum_{j,k=1}^n X_{ik}X_{kj}\theta_j\biggr)\biggr]\right)^2.\tag{3}$$

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

As a quick check that this is leading somewhere, for $n=2$ one has $Z_2=1+c_{11}+c_{22}+2c_{11}c_{22}$, giving $$\mathbb{E}[X^2+Y^2]=\int d\theta\int d\bar{\theta}\,\biggl(1+\bar{\theta}_{1}\theta_1+\bar{\theta}_{2}\theta_2+2\bar{\theta}_{1}\theta_1\bar{\theta}_{2}\theta_2\biggr)^2$$ $$\qquad=\int d\theta\int d\bar{\theta}\,\biggl(1+2\bar{\theta}_{1}\theta_1+2\bar{\theta}_{2}\theta_2+6\bar{\theta}_{1}\theta_1\bar{\theta}_{2}\theta_2\biggr)=6,\tag{4}$$ which is the correct answer.

Note that this also works for the multi-matrix generalization of Dan Piponi, $$\mathbb{E}\biggl[\sum_{i=1}^m X_m^2\biggr]=\int d\theta\int d\bar{\theta}\,\biggl(1+\bar{\theta}_{1}\theta_1+\bar{\theta}_{2}\theta_2+2\bar{\theta}_{1}\theta_1\bar{\theta}_{2}\theta_2\biggr)^m=m(m+1), $$ because only the term $\bar{\theta}_{1}\theta_1\bar{\theta}_{2}\theta_2$ gives a nonzero value upon integration over the Grassmann variables.

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 $Z_2=1+c_{11}+c_{22}+2c_{11}c_{22}$, 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.

Similarly, 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 $2m+m(m-1)=m(m+1)$.