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T. Amdeberhan
T. Amdeberhan
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For a vector $v$ of length $2^{n-2}$, denote by $v'$ the "mirrored" vector $J_{n-1}v$. If $v$ is an eigenvector of $B_{n-1}$ for the eigenvalue $\lambda$, then $B_n\binom v{\pm v'}=(\lambda\pm b_n)\binom v{\pm v'}$. So this gives you all the eigenvalues of $B_n$ as the sums $ b_1\pm\cdots\pm b_n$ (note that $ b_1$ never has a "$-$" sign) and the eigenvectors as a certain subset (respecting the dyadically recursive structure) of the set of all vectors $ (1,\pm1,...,\pm 1)$ where an even number of minus signs is used.

The determinant is: $\det(B_n)=\prod(b_1\pm b_2\pm\cdots\pm b_n)$.

For a vector $v$ of length $2^{n-2}$, denote by $v'$ the "mirrored" vector $J_{n-1}v$. If $v$ is an eigenvector of $B_{n-1}$ for the eigenvalue $\lambda$, then $B_n\binom v{\pm v'}=(\lambda\pm b_n)\binom v{\pm v'}$. So this gives you all the eigenvalues of $B_n$ as the sums $ b_1\pm\cdots\pm b_n$ (note that $ b_1$ never has a "$-$" sign) and the eigenvectors as a certain subset (respecting the dyadically recursive structure) of the set of all vectors $ (1,\pm1,...,\pm 1)$ where an even number of minus signs is used.

For a vector $v$ of length $2^{n-2}$, denote by $v'$ the "mirrored" vector $J_{n-1}v$. If $v$ is an eigenvector of $B_{n-1}$ for the eigenvalue $\lambda$, then $B_n\binom v{\pm v'}=(\lambda\pm b_n)\binom v{\pm v'}$. So this gives you all the eigenvalues of $B_n$ as the sums $ b_1\pm\cdots\pm b_n$ (note that $ b_1$ never has a "$-$" sign) and the eigenvectors as a certain subset (respecting the dyadically recursive structure) of the set of all vectors $ (1,\pm1,...,\pm 1)$ where an even number of minus signs is used.

The determinant is: $\det(B_n)=\prod(b_1\pm b_2\pm\cdots\pm b_n)$.

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Wolfgang
Wolfgang
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For a vector $v$ of length $2^{n-2}$, denote by $v'$ the "mirrored" vector $J_{n-1}v$. If $v$ is an eigenvector of $B_{n-1}$ for the eigenvalue $\lambda$, then $B_n\binom v{\pm v'}=(\lambda\pm b_n)\binom v{\pm v'}$. So this gives you all the eigenvalues of $B_n$ as the sums $ b_1\pm\cdots\pm b_n$ (note that $ b_1$ never has a "$-$" sign) and the eigenvectors as a certain subset (respecting the dyadically recursive structure) of the set of all vectors $ (\pm1,...,\pm 1)$$ (1,\pm1,...,\pm 1)$ where an even number of minus signs is used.

For a vector $v$ of length $2^{n-2}$, denote by $v'$ the "mirrored" vector $J_{n-1}v$. If $v$ is an eigenvector of $B_{n-1}$ for the eigenvalue $\lambda$, then $B_n\binom v{\pm v'}=(\lambda\pm b_n)\binom v{\pm v'}$. So this gives you all the eigenvalues of $B_n$ as the sums $ b_1\pm\cdots\pm b_n$ (note that $ b_1$ never has a "$-$" sign) and the eigenvectors as all vectors $ (\pm1,...,\pm 1)$ where an even number of minus signs is used.

For a vector $v$ of length $2^{n-2}$, denote by $v'$ the "mirrored" vector $J_{n-1}v$. If $v$ is an eigenvector of $B_{n-1}$ for the eigenvalue $\lambda$, then $B_n\binom v{\pm v'}=(\lambda\pm b_n)\binom v{\pm v'}$. So this gives you all the eigenvalues of $B_n$ as the sums $ b_1\pm\cdots\pm b_n$ (note that $ b_1$ never has a "$-$" sign) and the eigenvectors as a certain subset (respecting the dyadically recursive structure) of the set of all vectors $ (1,\pm1,...,\pm 1)$ where an even number of minus signs is used.

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Wolfgang
Wolfgang
  • 13.4k
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For a vector $v$ of length $2^{n-2}$, denote by $v'$ the "mirrored" vector $J_{n-1}v$. If $v$ is an eigenvector of $B_{n-1}$ for the eigenvalue $\lambda$, then $B_n\binom v{\pm v'}=(\lambda\pm b_n)\binom v{\pm v'}$. So this gives you all the eigenvalues of $B_n$ as the sums $ b_1\pm\cdots\pm b_n$ (note that $ b_1$ never has a "$-$" sign) and the eigenvectors as all vectors $ (\pm1,...,\pm 1)$ where an even number of minus signs is used.