Revisions to Kronecker product: Is it possible to simplify this product $e^{-A} \otimes e^{A}$ where $A$ is an invertible and symmetric matrix [closed]

Post Closed as "Not suitable for this site" by Steven Landsburg, Gro-Tsen, user44191, Roland Bacher, Brian Hopkins

occurred Oct 26, 2022 at 15:50

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edited Oct 19, 2022 at 8:28

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Let $A$ be an invertible, symmetric and tridiagonal matrix of size $n \times n$. Assume that $A_{i,i}=a \neq 0$ for $i=1\dotsc n$ and all the upper and lower diagonal elements in the sub- and super-diagonal of $A$ are $b \neq 0$. I would like to simplify the following Kronecker product: $e^{-A} \otimes e^{A}$.

I know that, given the Kronecker sum property of matrix exponential ($e^{A\oplus B}= e^{A}\otimes e^{B}$), the following holds:

\begin{equation} e^{-A} \otimes e^{A} = e^{-A \otimes I_n +I_n \otimes A}. \end{equation}

Since $A \otimes I_n$ and $I_n \otimes A$ commutes, using Zassenhaus formula,

\begin{equation} e^{-A} \otimes e^{A} = e^{-A \otimes I_n} e^{I_n \otimes A}=(e^{-A}\otimes I_n)(I_n \otimes e^{A}) \end{equation}

Given the above mentioned properties of matrix $A$, I was wondering whether it would be possible to further simplify this expression.

Let $A$ be an invertible, symmetric and tridiagonal matrix of size $n \times n$. Assume that $A_{i,i}=a \neq 0$ for $i=1\dotsc n$ and all the upper and lower diagonal elements of $A$ are $b \neq 0$. I would like to simplify the following Kronecker product: $e^{-A} \otimes e^{A}$.

I know that, given the Kronecker sum property of matrix exponential ($e^{A\oplus B}= e^{A}\otimes e^{B}$), the following holds:

\begin{equation} e^{-A} \otimes e^{A} = e^{-A \otimes I_n +I_n \otimes A}. \end{equation}

Since $A \otimes I_n$ and $I_n \otimes A$ commutes, using Zassenhaus formula,

\begin{equation} e^{-A} \otimes e^{A} = e^{-A \otimes I_n} e^{I_n \otimes A}=(e^{-A}\otimes I_n)(I_n \otimes e^{A}) \end{equation}

Given the above mentioned properties of matrix $A$, I was wondering whether it would be possible to further simplify this expression.

Let $A$ be an invertible, symmetric and tridiagonal matrix of size $n \times n$. Assume that $A_{i,i}=a \neq 0$ for $i=1\dotsc n$ and all the elements in the sub- and super-diagonal of $A$ are $b \neq 0$. I would like to simplify the following Kronecker product: $e^{-A} \otimes e^{A}$.

I know that, given the Kronecker sum property of matrix exponential ($e^{A\oplus B}= e^{A}\otimes e^{B}$), the following holds:

\begin{equation} e^{-A} \otimes e^{A} = e^{-A \otimes I_n +I_n \otimes A}. \end{equation}

Since $A \otimes I_n$ and $I_n \otimes A$ commutes, using Zassenhaus formula,

\begin{equation} e^{-A} \otimes e^{A} = e^{-A \otimes I_n} e^{I_n \otimes A}=(e^{-A}\otimes I_n)(I_n \otimes e^{A}) \end{equation}

Given the above mentioned properties of matrix $A$, I was wondering whether it would be possible to further simplify this expression.

Corrected errors in the equations as suggested in the comment section

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edited Oct 19, 2022 at 6:40

Mirar

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Let $A$ be an invertible, symmetric and tridiagonal matrix of size $n \times n$. Assume that $A_{i,i}=a \neq 0$ for $i=1\dotsc n$ and all the upper and lower diagonal elements of $A$ are $b \neq 0$. I would like to simplify the following Kronecker product: $e^{-A} \otimes e^{A}$.

I know that, given the Kronecker sum property of matrix exponential ($e^{A\oplus B}= e^{A}\otimes e^{B}$), the following holds:

\begin{equation} e^{-A} \otimes e^{A} = e^{A \otimes I_n -I_n \otimes A}. \end{equation}\begin{equation} e^{-A} \otimes e^{A} = e^{-A \otimes I_n +I_n \otimes A}. \end{equation}

Since $A \otimes I_n$ and $I_n \otimes A$ commutes, using Zassenhaus formula,

\begin{equation} e^{-A} \otimes e^{A} = e^{A \otimes I_n} e^{-I_n \otimes A}=(e^{A}\otimes I_n)(I_n \otimes e^{-A}) \end{equation}\begin{equation} e^{-A} \otimes e^{A} = e^{-A \otimes I_n} e^{I_n \otimes A}=(e^{-A}\otimes I_n)(I_n \otimes e^{A}) \end{equation}

Given the above mentioned properties of matrix $A$, I was wondering whether it would be possible to further simplify this expression.

Let $A$ be an invertible, symmetric and tridiagonal matrix of size $n \times n$. Assume that $A_{i,i}=a \neq 0$ for $i=1\dotsc n$ and all the upper and lower diagonal elements of $A$ are $b \neq 0$. I would like to simplify the following Kronecker product: $e^{-A} \otimes e^{A}$.

I know that, given the Kronecker sum property of matrix exponential ($e^{A\oplus B}= e^{A}\otimes e^{B}$), the following holds:

\begin{equation} e^{-A} \otimes e^{A} = e^{A \otimes I_n -I_n \otimes A}. \end{equation}

Since $A \otimes I_n$ and $I_n \otimes A$ commutes, using Zassenhaus formula,

\begin{equation} e^{-A} \otimes e^{A} = e^{A \otimes I_n} e^{-I_n \otimes A}=(e^{A}\otimes I_n)(I_n \otimes e^{-A}) \end{equation}

Given the above mentioned properties of matrix $A$, I was wondering whether it would be possible to further simplify this expression.

Let $A$ be an invertible, symmetric and tridiagonal matrix of size $n \times n$. Assume that $A_{i,i}=a \neq 0$ for $i=1\dotsc n$ and all the upper and lower diagonal elements of $A$ are $b \neq 0$. I would like to simplify the following Kronecker product: $e^{-A} \otimes e^{A}$.

I know that, given the Kronecker sum property of matrix exponential ($e^{A\oplus B}= e^{A}\otimes e^{B}$), the following holds:

\begin{equation} e^{-A} \otimes e^{A} = e^{-A \otimes I_n +I_n \otimes A}. \end{equation}

Since $A \otimes I_n$ and $I_n \otimes A$ commutes, using Zassenhaus formula,

\begin{equation} e^{-A} \otimes e^{A} = e^{-A \otimes I_n} e^{I_n \otimes A}=(e^{-A}\otimes I_n)(I_n \otimes e^{A}) \end{equation}

Given the above mentioned properties of matrix $A$, I was wondering whether it would be possible to further simplify this expression.

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edited Oct 18, 2022 at 21:57

Mirar

350
1
7

Let $A$ be an invertible, symmetric and triangulartridiagonal matrix of size $n \times n$. Assume that $A_{i,i}=a \neq 0$ for $i=1\dotsc n$ and all the upper and lower diagonal elements of $A$ are $b \neq 0$. I would like to simplify the following Kronecker product: $e^{-A} \otimes e^{A}$.

I know that, given the Kronecker sum property of matrix exponential ($e^{A\oplus B}= e^{A}\otimes e^{B}$), the following holds:

\begin{equation} e^{-A} \otimes e^{A} = e^{A \otimes I_n -I_n \otimes A}. \end{equation}

Since $A \otimes I_n$ and $I_n \otimes A$ commutes, using Zassenhaus formula,

\begin{equation} e^{-A} \otimes e^{A} = e^{A \otimes I_n} e^{-I_n \otimes A}=(e^{A}\otimes I_n)(I_n \otimes e^{-A}) \end{equation}

Given the above mentioned properties of matrix $A$, I was wondering whether it would be possible to further simplify this expression.

Let $A$ be an invertible, symmetric and triangular matrix of size $n \times n$. Assume that $A_{i,i}=a \neq 0$ for $i=1\dotsc n$ and all the upper and lower diagonal elements of $A$ are $b \neq 0$. I would like to simplify the following Kronecker product: $e^{-A} \otimes e^{A}$.