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Consider two column vectors $\mathbf{a}$ and $\mathbf{b}$ of length $k$ and $m$ respectively, $km$ variables denoted $y_{i,j}$ (i=1 to k, j=1 to m), and a quadratic form $\mathbf{y}^{T}\mathbb{M}\mathbf{y}$ where $$\mathbb{M}=\mathbb{D}_{1/a}\otimes\mathbb{D}_{1/b}+t(\mathbf{1}_{k}\mathbf{1}_{k}^{T}\otimes\mathbb{D}_{1/b}+\mathbb{D}_{1/a}\otimes\mathbf{1}_{m}\mathbf{1}_{m}^{T})$$with $\mathbb{D}_{1/a}$ being a $k$ by $k$ main-diagonal matrix with $1/a_i$ on the main digagonal and analogously with $\mathbb{D}_{1/b}$, while $\mathbf{1}_k$ is a column vector of length $k$ with all elements equal to $1$ (analogously for $\mathbf{1}_m$). Using results of my inquiry "Prove a special determinant formula" one can show that the determinant of $\mathbb{M}$ is

$$(1+t*\mathbf{b}^T\mathbf{1}_m)^{k-1}(1+t*\mathbf{a}^T\mathbf{1}_k)^{m-1}(1+t*(\mathbf{a}^T\mathbf{1}_k+\mathbf{b}^T\mathbf{1}_m))/(\prod{a_i})^{m}/(\prod{b_j })^k.$$

My new question is: suppose we replace $y_{k,m}$ (the last of our variables) by minus the sum of all the remaining $y_{i,j}$'s; we get a new quadratic form in the remaining $km-1$ variables. How does one prove that the determinant of the new ($km-1$ by $km-1$) matrix of coefficients is $$(1+t*\mathbf{b}^T\mathbf{1}_m)^{k-1}(1+t*\mathbf{a}^T\mathbf{1}_k)^{m-1}(\mathbf{a}^T\mathbf{1}_k)(\mathbf{b}^T\mathbf{1}_m)/(\prod{a_i})^{m}/(\prod{b_j })^k?$$

This has been posted on the Mathematics site for 3 months with no response; I hope someone can help me here.

To restate my question in strictly matrix language: Let $\mathbb{X}$ be a $km-1$ by $km$ matrix consisting of $km-1$ by $km-1$ identity matrix with an extra column whose all elements are equal to $-1$; find the determinant of $\mathbb{XMX^{T}}$.

The inverse of the last matrix is$$\mathbb{D}_{a}\otimes\mathbb{D}_{b}-t*\mathbf{a}\mathbf{a}^{T}\otimes\mathbb{D}_{b}/(1-t*\mathbf{a}^T\mathbf{1}_k)-t*\mathbb{D}_{a}\otimes\mathbf{b}\mathbf{b}^{T}/(1-t*\mathbf{b}^T\mathbf{1}_m)+\mathbf{a}\mathbf{a}^{T}\otimes\mathbf{b}\mathbf{b}^{T}(1-1/(1+t*\mathbf{a}\mathbf{a}^{T})-1/(1+t*\mathbf{b}\mathbf{b}^{T}))/(\mathbf{a}\mathbf{a}^{T}*\mathbf{b}\mathbf{b}^{T})$$after we delete its last row and the last column, but I do not know how to prove it either.

Consider two column vectors $\mathbf{a}$ and $\mathbf{b}$ of length $k$ and $m$ respectively, $km$ variables denoted $y_{i,j}$ (i=1 to k, j=1 to m), and a quadratic form $\mathbf{y}^{T}\mathbb{M}\mathbf{y}$ where $$\mathbb{M}=\mathbb{D}_{1/a}\otimes\mathbb{D}_{1/b}+t(\mathbf{1}_{k}\mathbf{1}_{k}^{T}\otimes\mathbb{D}_{1/b}+\mathbb{D}_{1/a}\otimes\mathbf{1}_{m}\mathbf{1}_{m}^{T})$$with $\mathbb{D}_{1/a}$ being a $k$ by $k$ main-diagonal matrix with $1/a_i$ on the main digagonal and analogously with $\mathbb{D}_{1/b}$, while $\mathbf{1}_k$ is a column vector of length $k$ with all elements equal to $1$ (analogously for $\mathbf{1}_m$). Using results of my inquiry "Prove a special determinant formula" one can show that the determinant of $\mathbb{M}$ is

$$(1+t*\mathbf{b}^T\mathbf{1}_m)^{k-1}(1+t*\mathbf{a}^T\mathbf{1}_k)^{m-1}(1+t*(\mathbf{a}^T\mathbf{1}_k+\mathbf{b}^T\mathbf{1}_m))/(\prod{a_i})^{m}/(\prod{b_j })^k.$$

My new question is: suppose we replace $y_{k,m}$ (the last of our variables) by minus the sum of all the remaining $y_{i,j}$'s; we get a new quadratic form in the remaining $km-1$ variables. How does one prove that the determinant of the new ($km-1$ by $km-1$) matrix of coefficients is $$(1+t*\mathbf{b}^T\mathbf{1}_m)^{k-1}(1+t*\mathbf{a}^T\mathbf{1}_k)^{m-1}(\mathbf{a}^T\mathbf{1}_k)(\mathbf{b}^T\mathbf{1}_m)/(\prod{a_i})^{m}/(\prod{b_j })^k?$$

This has been posted on the Mathematics site for 3 months with no response; I hope someone can help me here.

To restate my question in strictly matrix language: Let $\mathbb{X}$ be a $km-1$ by $km$ matrix consisting of $km-1$ by $km-1$ identity matrix with an extra column whose all elements are equal to $-1$; find the determinant of $\mathbb{XMX^{T}}$.

The inverse of the last matrix is$$\mathbb{D}_{a}\otimes\mathbb{D}_{b}-t*\mathbf{a}\mathbf{a}^{T}\otimes\mathbb{D}_{b}/(1-t*\mathbf{a}^T\mathbf{1}_k)-t*\mathbb{D}_{a}\otimes\mathbf{b}\mathbf{b}^{T}/(1-t*\mathbf{b}^T\mathbf{1}_m)+\mathbf{a}\mathbf{a}^{T}\otimes\mathbf{b}\mathbf{b}^{T}(1-1/(1+t*\mathbf{a}^T\mathbf{1}_k)-1/(1+t*\mathbf{b}^T\mathbf{1}_m))/(\mathbf{a}^T\mathbf{1}_k*\mathbf{b}^T\mathbf{1}_m)$$after we delete its last row and the last column, but I do not know how to prove it either.

Consider two column vectors $\mathbf{a}$ and $\mathbf{b}$ of length $k$ and $m$ respectively, $km$ variables denoted $y_{i,j}$ (i=1 to k, j=1 to m), and a quadratic form $\mathbf{y}^{T}\mathbb{M}\mathbf{y}$ where $$\mathbb{M}=\mathbb{D}_{1/a}\otimes\mathbb{D}_{1/b}+t(\mathbf{1}_{k}\mathbf{1}_{k}^{T}\otimes\mathbb{D}_{1/b}+\mathbb{D}_{1/a}\otimes\mathbf{1}_{m}\mathbf{1}_{m}^{T})$$with $\mathbb{D}_{1/a}$ being a $k$ by $k$ main-diagonal matrix with $1/a_i$ on the main digagonal and analogously with $\mathbb{D}_{1/b}$, while $\mathbf{1}_k$ is a column vector of length $k$ with all elements equal to $1$ (analogously for $\mathbf{1}_m$). Using results of my inquiry "Prove a special determinant formula" one can show that the determinant of $\mathbb{M}$ is

$$(1+t*\mathbf{b}^T\mathbf{1}_m)^{k-1}(1+t*\mathbf{a}^T\mathbf{1}_k)^{m-1}(1+t*(\mathbf{a}^T\mathbf{1}_k+\mathbf{b}^T\mathbf{1}_m))/(\prod{a_i})^{m}/(\prod{b_j })^k.$$

My new question is: suppose we replace $y_{k,m}$ (the last of our variables) by minus the sum of all the remaining $y_{i,j}$'s; we get a new quadratic form in the remaining $km-1$ variables. How does one prove that the determinant of the new ($km-1$ by $km-1$) matrix of coefficients is $$(1+t*\mathbf{b}^T\mathbf{1}_m)^{k-1}(1+t*\mathbf{a}^T\mathbf{1}_k)^{m-1}(\mathbf{a}^T\mathbf{1}_k)(\mathbf{b}^T\mathbf{1}_m)/(\prod{a_i})^{m}/(\prod{b_j })^k?$$

This has been posted on the Mathematics site for 3 months with no response; I hope someone can help me here.

To restate my question in strictly matrix language: Let $\mathbb{X}$ be a $km-1$ by $km$ matrix consisting of $km-1$ by $km-1$ identity matrix with an extra column whose all elements are equal to $-1$; find the determinant of $\mathbb{XMX^{T}}$.

Consider two column vectors $\mathbf{a}$ and $\mathbf{b}$ of length $k$ and $m$ respectively, $km$ variables denoted $y_{i,j}$ (i=1 to k, j=1 to m), and a quadratic form $\mathbf{y}^{T}\mathbb{M}\mathbf{y}$ where $$\mathbb{M}=\mathbb{D}_{1/a}\otimes\mathbb{D}_{1/b}+t(\mathbf{1}_{k}\mathbf{1}_{k}^{T}\otimes\mathbb{D}_{1/b}+\mathbb{D}_{1/a}\otimes\mathbf{1}_{m}\mathbf{1}_{m}^{T})$$with $\mathbb{D}_{1/a}$ being a $k$ by $k$ main-diagonal matrix with $1/a_i$ on the main digagonal and analogously with $\mathbb{D}_{1/b}$, while $\mathbf{1}_k$ is a column vector of length $k$ with all elements equal to $1$ (analogously for $\mathbf{1}_m$). Using results of my inquiry "Prove a special determinant formula" one can show that the determinant of $\mathbb{M}$ is

$$(1+t*\mathbf{b}^T\mathbf{1}_m)^{k-1}(1+t*\mathbf{a}^T\mathbf{1}_k)^{m-1}(1+t*(\mathbf{a}^T\mathbf{1}_k+\mathbf{b}^T\mathbf{1}_m))/(\prod{a_i})^{m}/(\prod{b_j })^k.$$

My new question is: suppose we replace $y_{k,m}$ (the last of our variables) by minus the sum of all the remaining $y_{i,j}$'s; we get a new quadratic form in the remaining $km-1$ variables. How does one prove that the determinant of the new ($km-1$ by $km-1$) matrix of coefficients is $$(1+t*\mathbf{b}^T\mathbf{1}_m)^{k-1}(1+t*\mathbf{a}^T\mathbf{1}_k)^{m-1}(\mathbf{a}^T\mathbf{1}_k)(\mathbf{b}^T\mathbf{1}_m)/(\prod{a_i})^{m}/(\prod{b_j })^k?$$

This has been posted on the Mathematics site for 3 months with no response; I hope someone can help me here.

To restate my question in strictly matrix language: Let $\mathbb{X}$ be a $km-1$ by $km$ matrix consisting of $km-1$ by $km-1$ identity matrix with an extra column whose all elements are equal to $-1$; find the determinant of $\mathbb{XMX^{T}}$.

The inverse of the last matrix is$$\mathbb{D}_{a}\otimes\mathbb{D}_{b}-t*\mathbf{a}\mathbf{a}^{T}\otimes\mathbb{D}_{b}/(1-t*\mathbf{a}^T\mathbf{1}_k)-t*\mathbb{D}_{a}\otimes\mathbf{b}\mathbf{b}^{T}/(1-t*\mathbf{b}^T\mathbf{1}_m)+\mathbf{a}\mathbf{a}^{T}\otimes\mathbf{b}\mathbf{b}^{T}(1-1/(1+t*\mathbf{a}\mathbf{a}^{T})-1/(1+t*\mathbf{b}\mathbf{b}^{T}))/(\mathbf{a}\mathbf{a}^{T}*\mathbf{b}\mathbf{b}^{T})$$after we delete its last row and the last column, but I do not know how to prove it either.

Consider two column vectors $\mathbf{a}$ and $\mathbf{b}$ of length $k$ and $m$ respectively, $km$ variables denoted $y_{i,j}$ (i=1 to k, j=1 to m), and a quadratic form $\mathbf{y}^{T}\mathbb{M}\mathbf{y}$ where $$\mathbb{M}=\mathbb{D}_{1/a}\otimes\mathbb{D}_{1/b}+t(\mathbf{1}_{k}\mathbf{1}_{k}^{T}\otimes\mathbb{D}_{1/b}+\mathbb{D}_{1/a}\otimes\mathbf{1}_{m}\mathbf{1}_{m}^{T})$$with $\mathbb{D}_{1/a}$ being a $k$ by $k$ main-diagonal matrix with $1/a_i$ on the main digagonal and analogously with $\mathbb{D}_{1/b}$, while $\mathbf{1}_k$ is a column vector of length $k$ with all elements equal to $1$ (analogously for $\mathbf{1}_m$). Using results of my inquiry "Prove a special determinant formula" one can show that the determinant of $\mathbb{M}$ is

$$(1+t*\mathbf{b}^T\mathbf{1}_m)^{k-1}(1+t*\mathbf{a}^T\mathbf{1}_k)^{m-1}(1+t*(\mathbf{a}^T\mathbf{1}_k+\mathbf{b}^T\mathbf{1}_m))/(\prod{a_i})^{m}/(\prod{b_j })^k.$$

My new question is: suppose we replace $y_{k,m}$ (the last of our variables) by minus the sum of all the remaining $y_{i,j}$'s; we get a new quadratic form in the remaining $km-1$ variables. How does one prove that the determinant of the new ($km-1$ by $km-1$) matrix of coefficients is $$(1+t*\mathbf{b}^T\mathbf{1}_m)^{k-1}(1+t*\mathbf{a}^T\mathbf{1}_k)^{m-1}(\mathbf{a}^T\mathbf{1}_k)(\mathbf{b}^T\mathbf{1}_m)/(\prod{a_i})^{m}/(\prod{b_j })^k?$$

This has been posted on the Mathematics site for 3 months with no response; I hope someone can help me here.

Consider two column vectors $\mathbf{a}$ and $\mathbf{b}$ of length $k$ and $m$ respectively, $km$ variables denoted $y_{i,j}$ (i=1 to k, j=1 to m), and a quadratic form $\mathbf{y}^{T}\mathbb{M}\mathbf{y}$ where $$\mathbb{M}=\mathbb{D}_{1/a}\otimes\mathbb{D}_{1/b}+t(\mathbf{1}_{k}\mathbf{1}_{k}^{T}\otimes\mathbb{D}_{1/b}+\mathbb{D}_{1/a}\otimes\mathbf{1}_{m}\mathbf{1}_{m}^{T})$$with $\mathbb{D}_{1/a}$ being a $k$ by $k$ main-diagonal matrix with $1/a_i$ on the main digagonal and analogously with $\mathbb{D}_{1/b}$, while $\mathbf{1}_k$ is a column vector of length $k$ with all elements equal to $1$ (analogously for $\mathbf{1}_m$). Using results of my inquiry "Prove a special determinant formula" one can show that the determinant of $\mathbb{M}$ is

$$(1+t*\mathbf{b}^T\mathbf{1}_m)^{k-1}(1+t*\mathbf{a}^T\mathbf{1}_k)^{m-1}(1+t*(\mathbf{a}^T\mathbf{1}_k+\mathbf{b}^T\mathbf{1}_m))/(\prod{a_i})^{m}/(\prod{b_j })^k.$$

My new question is: suppose we replace $y_{k,m}$ (the last of our variables) by minus the sum of all the remaining $y_{i,j}$'s; we get a new quadratic form in the remaining $km-1$ variables. How does one prove that the determinant of the new ($km-1$ by $km-1$) matrix of coefficients is $$(1+t*\mathbf{b}^T\mathbf{1}_m)^{k-1}(1+t*\mathbf{a}^T\mathbf{1}_k)^{m-1}(\mathbf{a}^T\mathbf{1}_k)(\mathbf{b}^T\mathbf{1}_m)/(\prod{a_i})^{m}/(\prod{b_j })^k?$$

This has been posted on the Mathematics site for 3 months with no response; I hope someone can help me here.

To restate my question in strictly matrix language: Let $\mathbb{X}$ be a $km-1$ by $km$ matrix consisting of $km-1$ by $km-1$ identity matrix with an extra column whose all elements are equal to $-1$; find the determinant of $\mathbb{XMX^{T}}$.