This is certainly a well established topic. Probably originating at the NIST (which would then have been called the National Bureau of Standards.)

A Weighing Matrix of weight $w$ and order $n$ is an $n \times n$ $0,\pm1$ matrix $A$ with $A^tA=wI_n.$ The case $w=n$ is a Hadamard Matrix with $A^tA=nI_n.$ Other than $n=1,2$ this requires $n=4m.$ It is an open question if they exist for every such $n.$ According to the linked article $668, 716, 892, 1004, 1132, 1244, 1388, 1436, 1676, 1772, 1916, 1948,$ and $1964$ are the only multiples of $4$ up to $2000$ for which such a design is unknown.

So $\sigma^2_n=\frac{\sigma^2}{n}$ for $n$ such that there is a Hadamard matrix of order $n.$

I would have guessed that for such an $n$ the optimal thing (in whatever sense) to do for $n-1$ is take a Hadamard matrix and delete a row and column. You showed that in some sense that is not optimal for $4-1=3.$

I really wonder what the case is for $n=8-1=7.$ The underlying design is a Fano Plane which seems as if it must be optimal (yet it isn't). For example: $$\left[ \begin {array}{ccccccc} -1&-1&1&-1&1&1&1\\ 1 &-1&-1&1&-1&1&1\\ 1&1&-1&-1&1&-1&1 \\ 1&1&1&-1&-1&1&-1\\ -1&1&1&1&-1& -1&1\\ 1&-1&1&1&1&-1&-1\\ -1&1&-1& 1&1&1&-1\end {array} \right] $$

Here $AA^t=7I-(J-I)$ is all $-1$ except $7$'s on the diagonal. The inverse of this is $\frac14I+\frac18(J-I).$

If the top left entry is changed to $0$ then the maximum entry on the diagonal is still $\frac14$ but in four of the seven diagonal positions one has $\frac{13}{64}.$ So this is, in some sense, even better.

Definitely better is making the entire main diagonal $0.$ Then the entire diagonal of $(AA^t)^{-1}$ is $\frac{155}{788}=.1967005076$

This is certainly a well established topic. Probably originating at the NIST (which would then have been called the National Bureau of Standards.) Here is a 1958 paper on weighing designs.

A Weighing Matrix of weight $w$ and order $n$ is an $n \times n$ $0,\pm1$ matrix $A$ with $A^tA=wI_n.$ The case $w=n$ is a Hadamard Matrix with $A^tA=nI_n.$ Other than $n=1,2$ this requires $n=4m.$ It is an open question if they exist for every such $n.$ According to the linked article $668, 716, 892, 1004, 1132, 1244, 1388, 1436, 1676, 1772, 1916, 1948,$ and $1964$ are the only multiples of $4$ up to $2000$ for which such a design is unknown.

So $\sigma^2_n=\frac{\sigma^2}{n}$ for $n$ such that there is a Hadamard matrix of order $n.$

I would have guessed that for such an $n$ the optimal thing (in whatever sense) to do for $n-1$ is take a Hadamard matrix and delete a row and column. You showed that in some sense that is not optimal for $4-1=3.$

I really wonder what the case is for $n=8-1=7.$ The underlying design is a Fano Plane which seems as if it must be optimal (yet it isn't). For example here is a circulant matrix: $$\left[ \begin {array}{ccccccc} -1&-1&1&-1&1&1&1\\ 1 &-1&-1&1&-1&1&1\\ 1&1&-1&-1&1&-1&1 \\ 1&1&1&-1&-1&1&-1\\ -1&1&1&1&-1& -1&1\\ 1&-1&1&1&1&-1&-1\\ -1&1&-1& 1&1&1&-1\end {array} \right] $$

Here $AA^t=7I-(J-I)$ is all $-1$ except $7$'s on the diagonal. The inverse of this is $\frac14I+\frac18(J-I).$

If the top left entry is changed to $0$ then the maximum entry on the diagonal is still $\frac14$ but in four of the seven diagonal positions one has $\frac{13}{64}=.203125.$ So this is, in some sense, even better.

Nicer in another way is the circulant matrix with first row 01011-10 with $AA^t=4I$

Definitely better is making the entire main diagonal $0.$ So first row $0-11-1111.$ Then the entire diagonal of $(AA^t)^{-1}$ is $\frac{155}{788}\approx .1967$

A systematic look at circulant matrices seems worthwhile (and has probably been done.)

This is certainly a well established topic. Probably originating at the NIST (which would then have been called the National Bureau of Standards.)

A Weighing Matrix of weight $w$ and order $n$ is an $n \times n$ $0,\pm1$ matrix $A$ with $A^tA=wI_n.$ The case $w=n$ is a Hadamard Matrix with $A^tA=nI_n.$ Other than $n=1,2$ this requires $n=4m.$ It is an open question if they exist for every such $n.$ According to the linked article $668, 716, 892, 1004, 1132, 1244, 1388, 1436, 1676, 1772, 1916, 1948,$ and $1964$ are the only multiples of $4$ up to $2000$ for which such a design is unknown.

So $\sigma^2_n=\frac{\sigma^2}{n}$ for $n$ such that there is a Hadamard matrix of order $n.$

I would have guessed that for such an $n$ the optimal thing (in whatever sense) to do for $n-1$ is take a Hadamard matrix and delete a row and column. You showed that in some sense that is not optimal for $4-1=3.$

I really wonder what the case is for $n=8-1=7.$ The underlying design is a Fano Plane which seems as if it must be optimal (yet it isn't). For example: $$\left[ \begin {array}{ccccccc} -1&-1&1&-1&1&1&1\\ 1 &-1&-1&1&-1&1&1\\ 1&1&-1&-1&1&-1&1 \\ 1&1&1&-1&-1&1&-1\\ -1&1&1&1&-1& -1&1\\ 1&-1&1&1&1&-1&-1\\ -1&1&-1& 1&1&1&-1\end {array} \right] $$

Here $AA^t=7I-(J-I)$ is all $-1$ except $7$'s on the diagonal. The inverse of this is $\frac14I+\frac18(J-I).$

If the top left entry is changed to $0$ then the maximum entry on the diagonal is still $\frac14$ but in four of the seven diagonal positions one has $\frac{13}{64}.$ So this is, in some sense, even betterI didn't experiment with going further than that.

This is certainly a well established topic. Probably originating at the NIST (which would then have been called the National Bureau of Standards.)

A Weighing Matrix of weight $w$ and order $n$ is an $n \times n$ $0,\pm1$ matrix $A$ with $A^tA=wI_n.$ The case $w=n$ is a Hadamard Matrix with $A^tA=nI_n.$ Other than $n=1,2$ this requires $n=4m.$ It is an open question if they exist for every such $n.$ According to the linked article $668, 716, 892, 1004, 1132, 1244, 1388, 1436, 1676, 1772, 1916, 1948,$ and $1964$ are the only multiples of $4$ up to $2000$ for which such a design is unknown.

So $\sigma^2_n=\frac{\sigma^2}{n}$ for $n$ such that there is a Hadamard matrix of order $n.$

I would have guessed that for such an $n$ the optimal thing (in whatever sense) to do for $n-1$ is take a Hadamard matrix and delete a row and column. You showed that in some sense that is not optimal for $4-1=3.$

I really wonder what the case is for $n=8-1=7.$ The underlying design is a Fano Plane which seems as if it must be optimal (yet it isn't). For example: $$\left[ \begin {array}{ccccccc} -1&-1&1&-1&1&1&1\\ 1 &-1&-1&1&-1&1&1\\ 1&1&-1&-1&1&-1&1 \\ 1&1&1&-1&-1&1&-1\\ -1&1&1&1&-1& -1&1\\ 1&-1&1&1&1&-1&-1\\ -1&1&-1& 1&1&1&-1\end {array} \right] $$

Here $AA^t=7I-(J-I)$ is all $-1$ except $7$'s on the diagonal. The inverse of this is $\frac14I+\frac18(J-I).$

If the top left entry is changed to $0$ then the maximum entry on the diagonal is still $\frac14$ but in four of the seven diagonal positions one has $\frac{13}{64}.$ So this is, in some sense, even better.

Definitely better is making the entire main diagonal $0.$ Then the entire diagonal of $(AA^t)^{-1}$ is $\frac{155}{788}=.1967005076$