Eigenvectors of the Hadamard matrix?Eigenvectors of a certain big upper triangular matrixEigenvectors and eigenvalues of a tridiagonal Toeplitz matrixEigenvectors and eigenvalues of nonsymmetric Tridiagonal matrixEigenvectors and eigenvalues of Tridiagonal matrix with varying diagonal elementsEquivalence of Hadamard Graph and Hadamard MatrixEigenvalues and eigenvectors of tridiagonal matricesStability of eigenvectors for diagonal perturbationsRotatable matrix, its eigenvalues and eigenvectors
Eigenvectors of the Hadamard matrix?
Eigenvectors of a certain big upper triangular matrixEigenvectors and eigenvalues of a tridiagonal Toeplitz matrixEigenvectors and eigenvalues of nonsymmetric Tridiagonal matrixEigenvectors and eigenvalues of Tridiagonal matrix with varying diagonal elementsEquivalence of Hadamard Graph and Hadamard MatrixEigenvalues and eigenvectors of tridiagonal matricesStability of eigenvectors for diagonal perturbationsRotatable matrix, its eigenvalues and eigenvectors
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What is known about the eigenvectors of the $2^n times 2^n$ Hadamard matrix defined recursively by $H_1=(1)$ and $$ H_N=beginpmatrixH_N/2 & H_N/2 \ H_N/2 & -H_N/2endpmatrix, $$ where $N=2^n$?
linear-algebra fourier-analysis eigenvalues eigenvector
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MCHMCH
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What is known about the eigenvectors of the $2^n times 2^n$ Hadamard matrix defined recursively by $H_1=(1)$ and $$ H_N=beginpmatrixH_N/2 & H_N/2 \ H_N/2 & -H_N/2endpmatrix, $$ where $N=2^n$?
linear-algebra fourier-analysis eigenvalues eigenvector
asked 9 hours ago
MCHMCH
3351 silver badge9 bronze badges
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What is known about the eigenvectors of the $2^n times 2^n$ Hadamard matrix defined recursively by $H_1=(1)$ and $$ H_N=beginpmatrixH_N/2 & H_N/2 \ H_N/2 & -H_N/2endpmatrix, $$ where $N=2^n$?
linear-algebra fourier-analysis eigenvalues eigenvector
asked 9 hours ago
MCHMCH
3351 silver badge9 bronze badges
$endgroup$
What is known about the eigenvectors of the $2^n times 2^n$ Hadamard matrix defined recursively by $H_1=(1)$ and $$ H_N=beginpmatrixH_N/2 & H_N/2 \ H_N/2 & -H_N/2endpmatrix, $$ where $N=2^n$?
linear-algebra fourier-analysis eigenvalues eigenvector
linear-algebra fourier-analysis eigenvalues eigenvector
asked 9 hours ago
MCHMCH
3351 silver badge9 bronze badges
asked 9 hours ago
MCHMCH
3351 silver badge9 bronze badges
asked 9 hours ago
MCHMCH
3351 silver badge9 bronze badges
asked 9 hours ago
MCHMCH
3351 silver badge9 bronze badges
asked 9 hours ago
MCHMCH
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1 Answer
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The $2^ntimes 2^n$ dimensional Hadamard matrices $H_n$ are also called Sylvester matrices or Walsh matrices. There are only two distinct eigenvalues $pm 2^n/2$, so the eigenvectors are not in general orthogonal. An orthogonal basis of eigenvectors is constructed recursively in A note on the eigenvectors of Hadamard matrices of order $2^n$ (1982) and in Some observations on eigenvectors of Hadamard matrices of order $2^n$ (1984).
answered 8 hours ago
Carlo BeenakkerCarlo Beenakker
89.3k9 gold badges222 silver badges331 bronze badges
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Aren't the eigenvectors easy to compute from the fact that $H_2^n = underbraceH_2 otimes H_2 otimes dots otimes H_2_text$n$ times$? You can diagonalize a Kronecker product factor by factor.
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– Federico Poloni
6 hours ago
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would that give you an orthogonal basis ? (as I understood the cited papers, that was the aim, to provide an efficient orthogonalization)
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– Carlo Beenakker
6 hours ago
1
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Yes, if I am not missing anything. If $H_2 = QDQ^*$, with $Q$ orthogonal (which exists since $H_2$ is symmetric), then $H_2^n = (Qotimes Q otimes dots otimes Q)(Dotimes D otimes dots otimes D) (Qotimes Q otimes dots otimes Q)^*$. I tested this quickly with Octave and it seems to work.
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– Federico Poloni
6 hours ago
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+1 ! ...........
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– Carlo Beenakker
6 hours ago
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1 Answer
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1 Answer
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$begingroup$
The $2^ntimes 2^n$ dimensional Hadamard matrices $H_n$ are also called Sylvester matrices or Walsh matrices. There are only two distinct eigenvalues $pm 2^n/2$, so the eigenvectors are not in general orthogonal. An orthogonal basis of eigenvectors is constructed recursively in A note on the eigenvectors of Hadamard matrices of order $2^n$ (1982) and in Some observations on eigenvectors of Hadamard matrices of order $2^n$ (1984).
answered 8 hours ago
Carlo BeenakkerCarlo Beenakker
89.3k9 gold badges222 silver badges331 bronze badges
$endgroup$
2
$begingroup$
Aren't the eigenvectors easy to compute from the fact that $H_2^n = underbraceH_2 otimes H_2 otimes dots otimes H_2_text$n$ times$? You can diagonalize a Kronecker product factor by factor.
$endgroup$
– Federico Poloni
6 hours ago
$begingroup$
would that give you an orthogonal basis ? (as I understood the cited papers, that was the aim, to provide an efficient orthogonalization)
$endgroup$
– Carlo Beenakker
6 hours ago
1
$begingroup$
Yes, if I am not missing anything. If $H_2 = QDQ^*$, with $Q$ orthogonal (which exists since $H_2$ is symmetric), then $H_2^n = (Qotimes Q otimes dots otimes Q)(Dotimes D otimes dots otimes D) (Qotimes Q otimes dots otimes Q)^*$. I tested this quickly with Octave and it seems to work.
$endgroup$
– Federico Poloni
6 hours ago
$begingroup$
+1 ! ...........
$endgroup$
– Carlo Beenakker
6 hours ago
add a comment
|
$begingroup$
The $2^ntimes 2^n$ dimensional Hadamard matrices $H_n$ are also called Sylvester matrices or Walsh matrices. There are only two distinct eigenvalues $pm 2^n/2$, so the eigenvectors are not in general orthogonal. An orthogonal basis of eigenvectors is constructed recursively in A note on the eigenvectors of Hadamard matrices of order $2^n$ (1982) and in Some observations on eigenvectors of Hadamard matrices of order $2^n$ (1984).
answered 8 hours ago
Carlo BeenakkerCarlo Beenakker
89.3k9 gold badges222 silver badges331 bronze badges
$endgroup$
2
$begingroup$
Aren't the eigenvectors easy to compute from the fact that $H_2^n = underbraceH_2 otimes H_2 otimes dots otimes H_2_text$n$ times$? You can diagonalize a Kronecker product factor by factor.
$endgroup$
– Federico Poloni
6 hours ago
$begingroup$
would that give you an orthogonal basis ? (as I understood the cited papers, that was the aim, to provide an efficient orthogonalization)
$endgroup$
– Carlo Beenakker
6 hours ago
1
$begingroup$
Yes, if I am not missing anything. If $H_2 = QDQ^*$, with $Q$ orthogonal (which exists since $H_2$ is symmetric), then $H_2^n = (Qotimes Q otimes dots otimes Q)(Dotimes D otimes dots otimes D) (Qotimes Q otimes dots otimes Q)^*$. I tested this quickly with Octave and it seems to work.
$endgroup$
– Federico Poloni
6 hours ago
$begingroup$
+1 ! ...........
$endgroup$
– Carlo Beenakker
6 hours ago
add a comment
|
$begingroup$
The $2^ntimes 2^n$ dimensional Hadamard matrices $H_n$ are also called Sylvester matrices or Walsh matrices. There are only two distinct eigenvalues $pm 2^n/2$, so the eigenvectors are not in general orthogonal. An orthogonal basis of eigenvectors is constructed recursively in A note on the eigenvectors of Hadamard matrices of order $2^n$ (1982) and in Some observations on eigenvectors of Hadamard matrices of order $2^n$ (1984).
answered 8 hours ago
Carlo BeenakkerCarlo Beenakker
89.3k9 gold badges222 silver badges331 bronze badges
$endgroup$
The $2^ntimes 2^n$ dimensional Hadamard matrices $H_n$ are also called Sylvester matrices or Walsh matrices. There are only two distinct eigenvalues $pm 2^n/2$, so the eigenvectors are not in general orthogonal. An orthogonal basis of eigenvectors is constructed recursively in A note on the eigenvectors of Hadamard matrices of order $2^n$ (1982) and in Some observations on eigenvectors of Hadamard matrices of order $2^n$ (1984).
answered 8 hours ago
Carlo BeenakkerCarlo Beenakker
89.3k9 gold badges222 silver badges331 bronze badges
edited 7 hours ago
answered 8 hours ago
Carlo BeenakkerCarlo Beenakker
89.3k9 gold badges222 silver badges331 bronze badges
answered 8 hours ago
Carlo BeenakkerCarlo Beenakker
89.3k9 gold badges222 silver badges331 bronze badges
answered 8 hours ago
Carlo BeenakkerCarlo Beenakker
89.3k9 gold badges222 silver badges331 bronze badges
89.3k9 gold badges222 silver badges331 bronze badges
2
$begingroup$
Aren't the eigenvectors easy to compute from the fact that $H_2^n = underbraceH_2 otimes H_2 otimes dots otimes H_2_text$n$ times$? You can diagonalize a Kronecker product factor by factor.
$endgroup$
– Federico Poloni
6 hours ago
$begingroup$
would that give you an orthogonal basis ? (as I understood the cited papers, that was the aim, to provide an efficient orthogonalization)
$endgroup$
– Carlo Beenakker
6 hours ago
1
$begingroup$
Yes, if I am not missing anything. If $H_2 = QDQ^*$, with $Q$ orthogonal (which exists since $H_2$ is symmetric), then $H_2^n = (Qotimes Q otimes dots otimes Q)(Dotimes D otimes dots otimes D) (Qotimes Q otimes dots otimes Q)^*$. I tested this quickly with Octave and it seems to work.
$endgroup$
– Federico Poloni
6 hours ago
$begingroup$
+1 ! ...........
$endgroup$
– Carlo Beenakker
6 hours ago
add a comment
|
2
$begingroup$
Aren't the eigenvectors easy to compute from the fact that $H_2^n = underbraceH_2 otimes H_2 otimes dots otimes H_2_text$n$ times$? You can diagonalize a Kronecker product factor by factor.
$endgroup$
– Federico Poloni
6 hours ago
$begingroup$
would that give you an orthogonal basis ? (as I understood the cited papers, that was the aim, to provide an efficient orthogonalization)
$endgroup$
– Carlo Beenakker
6 hours ago
1
$begingroup$
Yes, if I am not missing anything. If $H_2 = QDQ^*$, with $Q$ orthogonal (which exists since $H_2$ is symmetric), then $H_2^n = (Qotimes Q otimes dots otimes Q)(Dotimes D otimes dots otimes D) (Qotimes Q otimes dots otimes Q)^*$. I tested this quickly with Octave and it seems to work.
$endgroup$
– Federico Poloni
6 hours ago
$begingroup$
+1 ! ...........
$endgroup$
– Carlo Beenakker
6 hours ago
2
2
$begingroup$
Aren't the eigenvectors easy to compute from the fact that $H_2^n = underbraceH_2 otimes H_2 otimes dots otimes H_2_text$n$ times$? You can diagonalize a Kronecker product factor by factor.
$endgroup$
– Federico Poloni
6 hours ago
$begingroup$
Aren't the eigenvectors easy to compute from the fact that $H_2^n = underbraceH_2 otimes H_2 otimes dots otimes H_2_text$n$ times$? You can diagonalize a Kronecker product factor by factor.
$endgroup$
– Federico Poloni
6 hours ago
$begingroup$
would that give you an orthogonal basis ? (as I understood the cited papers, that was the aim, to provide an efficient orthogonalization)
$endgroup$
– Carlo Beenakker
6 hours ago
$begingroup$
would that give you an orthogonal basis ? (as I understood the cited papers, that was the aim, to provide an efficient orthogonalization)
$endgroup$
– Carlo Beenakker
6 hours ago
1
1
$begingroup$
Yes, if I am not missing anything. If $H_2 = QDQ^*$, with $Q$ orthogonal (which exists since $H_2$ is symmetric), then $H_2^n = (Qotimes Q otimes dots otimes Q)(Dotimes D otimes dots otimes D) (Qotimes Q otimes dots otimes Q)^*$. I tested this quickly with Octave and it seems to work.
$endgroup$
– Federico Poloni
6 hours ago
$begingroup$
Yes, if I am not missing anything. If $H_2 = QDQ^*$, with $Q$ orthogonal (which exists since $H_2$ is symmetric), then $H_2^n = (Qotimes Q otimes dots otimes Q)(Dotimes D otimes dots otimes D) (Qotimes Q otimes dots otimes Q)^*$. I tested this quickly with Octave and it seems to work.
$endgroup$
– Federico Poloni
6 hours ago
$begingroup$
+1 ! ...........
$endgroup$
– Carlo Beenakker
6 hours ago
$begingroup$
+1 ! ...........
$endgroup$
– Carlo Beenakker
6 hours ago
add a comment
|
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