Help with coding a matrixProblem with writting into a matrix in MathematicaHow to create a general format of a matrix with specified entries?How to verify the convexity of a function?Problem with a Positive Definite Kernel/MatrixDEigenvalues with Robin B.C. sign problemGenerating a matrixmatrix perturbation general formula with random valuesAntisymmetric Matrix Eigenvector NormalizationProblem subtracting matricesHow to compute eigenvalues of a large symbolic matrix?
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Help with coding a matrix
Problem with writting into a matrix in MathematicaHow to create a general format of a matrix with specified entries?How to verify the convexity of a function?Problem with a Positive Definite Kernel/MatrixDEigenvalues with Robin B.C. sign problemGenerating a matrixmatrix perturbation general formula with random valuesAntisymmetric Matrix Eigenvector NormalizationProblem subtracting matricesHow to compute eigenvalues of a large symbolic matrix?
$begingroup$
I have a $n times n$ matrix $A$ with a full set of eigenvalues $lambda$ including repetitions.
I want to create the following $i times i$ matrix:
$$left(sum_a=2^i (a-1) |a-1⟩⟨a| right) + sum_j=1^i d_j
sum_b=j^i |b⟩⟨b-j+1| $$
where $|1⟩,...,|i⟩$ is the standard basis and $d_j (lambda) = sum_c=1^n lambda_c^j$.
Any help with coding this matrix in Mathematica would be greatly appreciated.
matrix eigenvalues
asked 8 hours ago
jacobi16jacobi16
113
New contributor
jacobi16 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I have a $n times n$ matrix $A$ with a full set of eigenvalues $lambda$ including repetitions.
I want to create the following $i times i$ matrix:
$$left(sum_a=2^i (a-1) |a-1⟩⟨a| right) + sum_j=1^i d_j
sum_b=j^i |b⟩⟨b-j+1| $$
where $|1⟩,...,|i⟩$ is the standard basis and $d_j (lambda) = sum_c=1^n lambda_c^j$.
Any help with coding this matrix in Mathematica would be greatly appreciated.
matrix eigenvalues
asked 8 hours ago
jacobi16jacobi16
113
New contributor
jacobi16 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
You must give an explicit mathematical formula for $|b rangle langle b-j+1|$ and other terms.
$endgroup$
– David G. Stork
8 hours ago
$begingroup$
@DavidG.Stork I added a missing definition and believe I have defined all the other terms. Please let me know if any definition is unclear.
$endgroup$
– jacobi16
8 hours ago
$begingroup$
The first term $delta_i,0$ is a scalar, not a matrix, and does not fit into the formula. Does it mean that for $i=0$ you want to get the scalar 1 as the answer?
$endgroup$
– Roman
7 hours ago
$begingroup$
Does $d_j(lambda)$ depend on $j$ at all?
$endgroup$
– Roman
7 hours ago
$begingroup$
@Roman Thanks for your comments. I had mistakenly written the superscript of $x$ as $q$ instead of $j$. $delta_i,0$ would be the identity matrix when $i=0$, which makes this term irrelevant now that I think about it. Sorry for the mistake with the definition of $d_j$ and thanks for spotting out the unnecessary $delta_i,0$ factor. I have edited my question to reflect these changes.
$endgroup$
– jacobi16
2 hours ago
add a comment |
$begingroup$
I have a $n times n$ matrix $A$ with a full set of eigenvalues $lambda$ including repetitions.
I want to create the following $i times i$ matrix:
$$left(sum_a=2^i (a-1) |a-1⟩⟨a| right) + sum_j=1^i d_j
sum_b=j^i |b⟩⟨b-j+1| $$
where $|1⟩,...,|i⟩$ is the standard basis and $d_j (lambda) = sum_c=1^n lambda_c^j$.
Any help with coding this matrix in Mathematica would be greatly appreciated.
matrix eigenvalues
asked 8 hours ago
jacobi16jacobi16
113
New contributor
jacobi16 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have a $n times n$ matrix $A$ with a full set of eigenvalues $lambda$ including repetitions.
I want to create the following $i times i$ matrix:
$$left(sum_a=2^i (a-1) |a-1⟩⟨a| right) + sum_j=1^i d_j
sum_b=j^i |b⟩⟨b-j+1| $$
where $|1⟩,...,|i⟩$ is the standard basis and $d_j (lambda) = sum_c=1^n lambda_c^j$.
Any help with coding this matrix in Mathematica would be greatly appreciated.
matrix eigenvalues
matrix eigenvalues
asked 8 hours ago
jacobi16jacobi16
113
New contributor
jacobi16 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
jacobi16jacobi16
113
New contributor
jacobi16 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
jacobi16
asked 8 hours ago
jacobi16jacobi16
113
New contributor
jacobi16 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
jacobi16jacobi16
113
asked 8 hours ago
jacobi16jacobi16
113
113
New contributor
jacobi16 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
jacobi16 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
You must give an explicit mathematical formula for $|b rangle langle b-j+1|$ and other terms.
$endgroup$
– David G. Stork
8 hours ago
$begingroup$
@DavidG.Stork I added a missing definition and believe I have defined all the other terms. Please let me know if any definition is unclear.
$endgroup$
– jacobi16
8 hours ago
$begingroup$
The first term $delta_i,0$ is a scalar, not a matrix, and does not fit into the formula. Does it mean that for $i=0$ you want to get the scalar 1 as the answer?
$endgroup$
– Roman
7 hours ago
$begingroup$
Does $d_j(lambda)$ depend on $j$ at all?
$endgroup$
– Roman
7 hours ago
$begingroup$
@Roman Thanks for your comments. I had mistakenly written the superscript of $x$ as $q$ instead of $j$. $delta_i,0$ would be the identity matrix when $i=0$, which makes this term irrelevant now that I think about it. Sorry for the mistake with the definition of $d_j$ and thanks for spotting out the unnecessary $delta_i,0$ factor. I have edited my question to reflect these changes.
$endgroup$
– jacobi16
2 hours ago
add a comment |
$begingroup$
You must give an explicit mathematical formula for $|b rangle langle b-j+1|$ and other terms.
$endgroup$
– David G. Stork
8 hours ago
$begingroup$
@DavidG.Stork I added a missing definition and believe I have defined all the other terms. Please let me know if any definition is unclear.
$endgroup$
– jacobi16
8 hours ago
$begingroup$
The first term $delta_i,0$ is a scalar, not a matrix, and does not fit into the formula. Does it mean that for $i=0$ you want to get the scalar 1 as the answer?
$endgroup$
– Roman
7 hours ago
$begingroup$
Does $d_j(lambda)$ depend on $j$ at all?
$endgroup$
– Roman
7 hours ago
$begingroup$
@Roman Thanks for your comments. I had mistakenly written the superscript of $x$ as $q$ instead of $j$. $delta_i,0$ would be the identity matrix when $i=0$, which makes this term irrelevant now that I think about it. Sorry for the mistake with the definition of $d_j$ and thanks for spotting out the unnecessary $delta_i,0$ factor. I have edited my question to reflect these changes.
$endgroup$
– jacobi16
2 hours ago
$begingroup$
You must give an explicit mathematical formula for $|b rangle langle b-j+1|$ and other terms.
$endgroup$
– David G. Stork
8 hours ago
$begingroup$
You must give an explicit mathematical formula for $|b rangle langle b-j+1|$ and other terms.
$endgroup$
– David G. Stork
8 hours ago
$begingroup$
@DavidG.Stork I added a missing definition and believe I have defined all the other terms. Please let me know if any definition is unclear.
$endgroup$
– jacobi16
8 hours ago
$begingroup$
@DavidG.Stork I added a missing definition and believe I have defined all the other terms. Please let me know if any definition is unclear.
$endgroup$
– jacobi16
8 hours ago
$begingroup$
The first term $delta_i,0$ is a scalar, not a matrix, and does not fit into the formula. Does it mean that for $i=0$ you want to get the scalar 1 as the answer?
$endgroup$
– Roman
7 hours ago
$begingroup$
The first term $delta_i,0$ is a scalar, not a matrix, and does not fit into the formula. Does it mean that for $i=0$ you want to get the scalar 1 as the answer?
$endgroup$
– Roman
7 hours ago
$begingroup$
Does $d_j(lambda)$ depend on $j$ at all?
$endgroup$
– Roman
7 hours ago
$begingroup$
Does $d_j(lambda)$ depend on $j$ at all?
$endgroup$
– Roman
7 hours ago
$begingroup$
@Roman Thanks for your comments. I had mistakenly written the superscript of $x$ as $q$ instead of $j$. $delta_i,0$ would be the identity matrix when $i=0$, which makes this term irrelevant now that I think about it. Sorry for the mistake with the definition of $d_j$ and thanks for spotting out the unnecessary $delta_i,0$ factor. I have edited my question to reflect these changes.
$endgroup$
– jacobi16
2 hours ago
$begingroup$
@Roman Thanks for your comments. I had mistakenly written the superscript of $x$ as $q$ instead of $j$. $delta_i,0$ would be the identity matrix when $i=0$, which makes this term irrelevant now that I think about it. Sorry for the mistake with the definition of $d_j$ and thanks for spotting out the unnecessary $delta_i,0$ factor. I have edited my question to reflect these changes.
$endgroup$
– jacobi16
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can also use a combination of ToeplitzMatrix, DiagonalMatrix, LowerTriangularize and SparseArray:
ClearAll[mat]
mat[n_] := Module[dd = Array[d, n],
LowerTriangularize[ToeplitzMatrix[dd, SparseArray]] +
DiagonalMatrix[SparseArray@Range[n - 1], 1]]
mat[5] // MatrixForm // TeXForm
$left(
beginarrayccccc
d(1) & 1 & 0 & 0 & 0 \
d(2) & d(1) & 2 & 0 & 0 \
d(3) & d(2) & d(1) & 3 & 0 \
d(4) & d(3) & d(2) & d(1) & 4 \
d(5) & d(4) & d(3) & d(2) & d(1) \
endarray
right)$
answered 6 hours ago
kglrkglr
199k10223451
$endgroup$
add a comment |
$begingroup$
You can define the matrix with
M[i_Integer?Positive] := SparseArray[Band[1, 2] -> Range[i - 1],
a_, b_ /; a >= b -> d[a - b + 1],
i, i]
I don't know what to do with the first term $delta_i,0$ because it is not an operator/matrix (it is a scalar).
Also, I don't understand your definition of $d_j=$d[j]: should it depend on $j$?
M[5] // MatrixForm
answered 7 hours ago
RomanRoman
11.3k11944
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can also use a combination of ToeplitzMatrix, DiagonalMatrix, LowerTriangularize and SparseArray:
ClearAll[mat]
mat[n_] := Module[dd = Array[d, n],
LowerTriangularize[ToeplitzMatrix[dd, SparseArray]] +
DiagonalMatrix[SparseArray@Range[n - 1], 1]]
mat[5] // MatrixForm // TeXForm
$left(
beginarrayccccc
d(1) & 1 & 0 & 0 & 0 \
d(2) & d(1) & 2 & 0 & 0 \
d(3) & d(2) & d(1) & 3 & 0 \
d(4) & d(3) & d(2) & d(1) & 4 \
d(5) & d(4) & d(3) & d(2) & d(1) \
endarray
right)$
answered 6 hours ago
kglrkglr
199k10223451
$endgroup$
add a comment |
$begingroup$
You can also use a combination of ToeplitzMatrix, DiagonalMatrix, LowerTriangularize and SparseArray:
ClearAll[mat]
mat[n_] := Module[dd = Array[d, n],
LowerTriangularize[ToeplitzMatrix[dd, SparseArray]] +
DiagonalMatrix[SparseArray@Range[n - 1], 1]]
mat[5] // MatrixForm // TeXForm
$left(
beginarrayccccc
d(1) & 1 & 0 & 0 & 0 \
d(2) & d(1) & 2 & 0 & 0 \
d(3) & d(2) & d(1) & 3 & 0 \
d(4) & d(3) & d(2) & d(1) & 4 \
d(5) & d(4) & d(3) & d(2) & d(1) \
endarray
right)$
answered 6 hours ago
kglrkglr
199k10223451
$endgroup$
add a comment |
$begingroup$
You can also use a combination of ToeplitzMatrix, DiagonalMatrix, LowerTriangularize and SparseArray:
ClearAll[mat]
mat[n_] := Module[dd = Array[d, n],
LowerTriangularize[ToeplitzMatrix[dd, SparseArray]] +
DiagonalMatrix[SparseArray@Range[n - 1], 1]]
mat[5] // MatrixForm // TeXForm
$left(
beginarrayccccc
d(1) & 1 & 0 & 0 & 0 \
d(2) & d(1) & 2 & 0 & 0 \
d(3) & d(2) & d(1) & 3 & 0 \
d(4) & d(3) & d(2) & d(1) & 4 \
d(5) & d(4) & d(3) & d(2) & d(1) \
endarray
right)$
answered 6 hours ago
kglrkglr
199k10223451
$endgroup$
You can also use a combination of ToeplitzMatrix, DiagonalMatrix, LowerTriangularize and SparseArray:
ClearAll[mat]
mat[n_] := Module[dd = Array[d, n],
LowerTriangularize[ToeplitzMatrix[dd, SparseArray]] +
DiagonalMatrix[SparseArray@Range[n - 1], 1]]
mat[5] // MatrixForm // TeXForm
$left(
beginarrayccccc
d(1) & 1 & 0 & 0 & 0 \
d(2) & d(1) & 2 & 0 & 0 \
d(3) & d(2) & d(1) & 3 & 0 \
d(4) & d(3) & d(2) & d(1) & 4 \
d(5) & d(4) & d(3) & d(2) & d(1) \
endarray
right)$
answered 6 hours ago
kglrkglr
199k10223451
answered 6 hours ago
kglrkglr
199k10223451
answered 6 hours ago
kglrkglr
199k10223451
answered 6 hours ago
kglrkglr
199k10223451
199k10223451
add a comment |
add a comment |
$begingroup$
You can define the matrix with
M[i_Integer?Positive] := SparseArray[Band[1, 2] -> Range[i - 1],
a_, b_ /; a >= b -> d[a - b + 1],
i, i]
I don't know what to do with the first term $delta_i,0$ because it is not an operator/matrix (it is a scalar).
Also, I don't understand your definition of $d_j=$d[j]: should it depend on $j$?
M[5] // MatrixForm
answered 7 hours ago
RomanRoman
11.3k11944
$endgroup$
add a comment |
$begingroup$
You can define the matrix with
M[i_Integer?Positive] := SparseArray[Band[1, 2] -> Range[i - 1],
a_, b_ /; a >= b -> d[a - b + 1],
i, i]
I don't know what to do with the first term $delta_i,0$ because it is not an operator/matrix (it is a scalar).
Also, I don't understand your definition of $d_j=$d[j]: should it depend on $j$?
M[5] // MatrixForm
answered 7 hours ago
RomanRoman
11.3k11944
$endgroup$
add a comment |
$begingroup$
You can define the matrix with
M[i_Integer?Positive] := SparseArray[Band[1, 2] -> Range[i - 1],
a_, b_ /; a >= b -> d[a - b + 1],
i, i]
I don't know what to do with the first term $delta_i,0$ because it is not an operator/matrix (it is a scalar).
Also, I don't understand your definition of $d_j=$d[j]: should it depend on $j$?
M[5] // MatrixForm
answered 7 hours ago
RomanRoman
11.3k11944
$endgroup$
You can define the matrix with
M[i_Integer?Positive] := SparseArray[Band[1, 2] -> Range[i - 1],
a_, b_ /; a >= b -> d[a - b + 1],
i, i]
I don't know what to do with the first term $delta_i,0$ because it is not an operator/matrix (it is a scalar).
Also, I don't understand your definition of $d_j=$d[j]: should it depend on $j$?
M[5] // MatrixForm
answered 7 hours ago
RomanRoman
11.3k11944
answered 7 hours ago
RomanRoman
11.3k11944
answered 7 hours ago
RomanRoman
11.3k11944
answered 7 hours ago
RomanRoman
11.3k11944
11.3k11944
add a comment |
add a comment |
jacobi16 is a new contributor. Be nice, and check out our Code of Conduct.
jacobi16 is a new contributor. Be nice, and check out our Code of Conduct.
jacobi16 is a new contributor. Be nice, and check out our Code of Conduct.
jacobi16 is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
You must give an explicit mathematical formula for $|b rangle langle b-j+1|$ and other terms.
$endgroup$
– David G. Stork
8 hours ago
$begingroup$
@DavidG.Stork I added a missing definition and believe I have defined all the other terms. Please let me know if any definition is unclear.
$endgroup$
– jacobi16
8 hours ago
$begingroup$
The first term $delta_i,0$ is a scalar, not a matrix, and does not fit into the formula. Does it mean that for $i=0$ you want to get the scalar 1 as the answer?
$endgroup$
– Roman
7 hours ago
$begingroup$
Does $d_j(lambda)$ depend on $j$ at all?
$endgroup$
– Roman
7 hours ago
$begingroup$
@Roman Thanks for your comments. I had mistakenly written the superscript of $x$ as $q$ instead of $j$. $delta_i,0$ would be the identity matrix when $i=0$, which makes this term irrelevant now that I think about it. Sorry for the mistake with the definition of $d_j$ and thanks for spotting out the unnecessary $delta_i,0$ factor. I have edited my question to reflect these changes.
$endgroup$
– jacobi16
2 hours ago