Need to find matrix formulation

Clash Royale CLAN TAG#URR8PPP

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I have a $B$ matrix: $B = B_ij$

I need to find closed matrix formulation of:

$$sum_i sum_j sum_m sum_n B_ij B_jm B_mn B_ni$$

But I am so confused!

Edit by Henrik:

Originally, it was asked to express

$$sum_i sum_j sum_m sum_n B_mi B_mj B_ni B_nj$$

in terms of matrices.

edited 25 mins ago

Henrik Schumacher

38.8k254114

asked 7 hours ago

farhadpti

New contributor

I formatted the math using Mathjax. Please check if I introduced any errors.
â€“Â Johu
7 hours ago

This site is about Mathematica, the software. Should we move your question over to Mathematics?
â€“Â Johu
7 hours ago

Could you also check, if you have not made any mistakes with the indexes.
â€“Â Johu
7 hours ago

@DavidG.Stork I am not sure whether your edit was meaningfull. By swapping some of the indices, you actually changed the question dramatically. For example, Louis' (imho correct) answer became invalid under this modification.
â€“Â Henrik Schumacher
31 mins ago

@LouisB I'd suggest to undelete your post. Imho, it is a perfect answer to the original question.
â€“Â Henrik Schumacher
19 mins ago

add a commentÂ |Â

up vote
1
down vote

favorite

I have a $B$ matrix: $B = B_ij$

I need to find closed matrix formulation of:

$$sum_i sum_j sum_m sum_n B_ij B_jm B_mn B_ni$$

But I am so confused!

Edit by Henrik:

Originally, it was asked to express

$$sum_i sum_j sum_m sum_n B_mi B_mj B_ni B_nj$$

in terms of matrices.

edited 25 mins ago

Henrik Schumacher

38.8k254114

asked 7 hours ago

farhadpti

New contributor

I formatted the math using Mathjax. Please check if I introduced any errors.
â€“Â Johu
7 hours ago

This site is about Mathematica, the software. Should we move your question over to Mathematics?
â€“Â Johu
7 hours ago

Could you also check, if you have not made any mistakes with the indexes.
â€“Â Johu
7 hours ago

@DavidG.Stork I am not sure whether your edit was meaningfull. By swapping some of the indices, you actually changed the question dramatically. For example, Louis' (imho correct) answer became invalid under this modification.
â€“Â Henrik Schumacher
31 mins ago

@LouisB I'd suggest to undelete your post. Imho, it is a perfect answer to the original question.
â€“Â Henrik Schumacher
19 mins ago

add a commentÂ |Â

up vote
1
down vote

favorite

I have a $B$ matrix: $B = B_ij$

I need to find closed matrix formulation of:

$$sum_i sum_j sum_m sum_n B_ij B_jm B_mn B_ni$$

But I am so confused!

Edit by Henrik:

Originally, it was asked to express

$$sum_i sum_j sum_m sum_n B_mi B_mj B_ni B_nj$$

in terms of matrices.

edited 25 mins ago

Henrik Schumacher

38.8k254114

asked 7 hours ago

farhadpti

New contributor

I have a $B$ matrix: $B = B_ij$

I need to find closed matrix formulation of:

$$sum_i sum_j sum_m sum_n B_ij B_jm B_mn B_ni$$

But I am so confused!

Edit by Henrik:

Originally, it was asked to express

$$sum_i sum_j sum_m sum_n B_mi B_mj B_ni B_nj$$

in terms of matrices.

matrix algebra

edited 25 mins ago

Henrik Schumacher

38.8k254114

asked 7 hours ago

farhadpti

New contributor

edited 25 mins ago

Henrik Schumacher

38.8k254114

asked 7 hours ago

farhadpti

New contributor

edited 25 mins ago

Henrik Schumacher

38.8k254114

edited 25 mins ago

Henrik Schumacher

38.8k254114

edited 25 mins ago

Henrik Schumacher

38.8k254114

asked 7 hours ago

farhadpti

New contributor

asked 7 hours ago

farhadpti

asked 7 hours ago

farhadpti

New contributor

farhadpti is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

I formatted the math using Mathjax. Please check if I introduced any errors.
â€“Â Johu
7 hours ago

This site is about Mathematica, the software. Should we move your question over to Mathematics?
â€“Â Johu
7 hours ago

Could you also check, if you have not made any mistakes with the indexes.
â€“Â Johu
7 hours ago

@DavidG.Stork I am not sure whether your edit was meaningfull. By swapping some of the indices, you actually changed the question dramatically. For example, Louis' (imho correct) answer became invalid under this modification.
â€“Â Henrik Schumacher
31 mins ago

@LouisB I'd suggest to undelete your post. Imho, it is a perfect answer to the original question.
â€“Â Henrik Schumacher
19 mins ago

add a commentÂ |Â

I formatted the math using Mathjax. Please check if I introduced any errors.
â€“Â Johu
7 hours ago

This site is about Mathematica, the software. Should we move your question over to Mathematics?
â€“Â Johu
7 hours ago

Could you also check, if you have not made any mistakes with the indexes.
â€“Â Johu
7 hours ago

@DavidG.Stork I am not sure whether your edit was meaningfull. By swapping some of the indices, you actually changed the question dramatically. For example, Louis' (imho correct) answer became invalid under this modification.
â€“Â Henrik Schumacher
31 mins ago

@LouisB I'd suggest to undelete your post. Imho, it is a perfect answer to the original question.
â€“Â Henrik Schumacher
19 mins ago

I formatted the math using Mathjax. Please check if I introduced any errors.
â€“Â Johu
7 hours ago

This site is about Mathematica, the software. Should we move your question over to Mathematics?
â€“Â Johu
7 hours ago

Could you also check, if you have not made any mistakes with the indexes.
â€“Â Johu
7 hours ago

@DavidG.Stork I am not sure whether your edit was meaningfull. By swapping some of the indices, you actually changed the question dramatically. For example, Louis' (imho correct) answer became invalid under this modification.
â€“Â Henrik Schumacher
31 mins ago

@LouisB I'd suggest to undelete your post. Imho, it is a perfect answer to the original question.
â€“Â Henrik Schumacher
19 mins ago

add a commentÂ |Â

3 Answers
3

active

oldest

votes

up vote
4
down vote

If this is indeed a Mathematica question, then first note that:

$$sum _j B_i,j B_j,kequiv (B.B)_i,j$$

and

$$sum _i B_i,iequiv operatornameTr[B]$$

So, the Mathematica equivalent of:

$$sum _i sum _j sum _m sum _n B_i,j B_j,m B_m,n B_n,i $$

is:

Tr[B . B . B . B]

or:

Tr[MatrixPower[B, 4]]

For the original form of the question, note that:

$$ B_i,jequiv B^T_j,i $$

So, the Mathematica equivalent of:

$$sum _i sum _j sum _k sum _l B_m,i B_m,j B_n,i B_n,j$$

is:

Tr[B.Transpose[B].B.Transpose[B]]

answered 5 hours ago

Carl Woll

58.1k274150

add a commentÂ |Â

up vote
1
down vote

This is a closed matrix formulation of $B^4$.

answered 7 hours ago

David G. Stork

21.5k11646

2

It's the trace of $B^4$, no?
â€“Â LouisB
6 hours ago

add a commentÂ |Â

up vote
0
down vote

Carl's answer is perfect as it is. His use of MatrixPower inspired me to think a little about computational complexity: Matrix-matrix multiplication has complexity $O(n^3)$ (in the straight-forward implementations, not these theoretically fancy but practically irrelevant algorithms). If we can get rid of one or two of them, we can even afford a transposition to speed up the computation:

n = 5000;
B = RandomReal[-1, 1, n, n];
a = Tr[B.B.B.B]; // RepeatedTiming // First
b = Tr[MatrixPower[B, 4]]; // RepeatedTiming // First
c = With[A = B.B, Total[A Transpose[A], 2]]; // RepeatedTiming // First
a == b == c

5.7

4.207

2.5

True

For the original question, LouisB's answer (c below) seems to be both correct and efficient:

n = 5000;
B = RandomReal[-1, 1, n, n];
a = Tr[B.Transpose[B].B.Transpose[B]]; // RepeatedTiming // First
b = With[u = Flatten[B.Transpose[B]], u.u]; // RepeatedTiming // First
c = Total[(Transpose[B].B)^2, 2]; // RepeatedTiming // First
a == b == c

5.8

2.4

2.40

True

edited 21 mins ago

answered 34 mins ago

Henrik Schumacher

38.8k254114

add a commentÂ |Â

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
4
down vote

If this is indeed a Mathematica question, then first note that:

$$sum _j B_i,j B_j,kequiv (B.B)_i,j$$

and

$$sum _i B_i,iequiv operatornameTr[B]$$

So, the Mathematica equivalent of:

$$sum _i sum _j sum _m sum _n B_i,j B_j,m B_m,n B_n,i $$

is:

Tr[B . B . B . B]

or:

Tr[MatrixPower[B, 4]]

For the original form of the question, note that:

$$ B_i,jequiv B^T_j,i $$

So, the Mathematica equivalent of:

$$sum _i sum _j sum _k sum _l B_m,i B_m,j B_n,i B_n,j$$

is:

Tr[B.Transpose[B].B.Transpose[B]]

answered 5 hours ago

Carl Woll

58.1k274150

add a commentÂ |Â

up vote
4
down vote

If this is indeed a Mathematica question, then first note that:

$$sum _j B_i,j B_j,kequiv (B.B)_i,j$$

and

$$sum _i B_i,iequiv operatornameTr[B]$$

So, the Mathematica equivalent of:

$$sum _i sum _j sum _m sum _n B_i,j B_j,m B_m,n B_n,i $$

is:

Tr[B . B . B . B]

or:

Tr[MatrixPower[B, 4]]

For the original form of the question, note that:

$$ B_i,jequiv B^T_j,i $$

So, the Mathematica equivalent of:

$$sum _i sum _j sum _k sum _l B_m,i B_m,j B_n,i B_n,j$$

is:

Tr[B.Transpose[B].B.Transpose[B]]

answered 5 hours ago

Carl Woll

58.1k274150

add a commentÂ |Â

up vote
4
down vote

If this is indeed a Mathematica question, then first note that:

$$sum _j B_i,j B_j,kequiv (B.B)_i,j$$

and

$$sum _i B_i,iequiv operatornameTr[B]$$

So, the Mathematica equivalent of:

$$sum _i sum _j sum _m sum _n B_i,j B_j,m B_m,n B_n,i $$

is:

Tr[B . B . B . B]

or:

Tr[MatrixPower[B, 4]]

For the original form of the question, note that:

$$ B_i,jequiv B^T_j,i $$

So, the Mathematica equivalent of:

$$sum _i sum _j sum _k sum _l B_m,i B_m,j B_n,i B_n,j$$

is:

Tr[B.Transpose[B].B.Transpose[B]]

answered 5 hours ago

Carl Woll

58.1k274150

If this is indeed a Mathematica question, then first note that:

$$sum _j B_i,j B_j,kequiv (B.B)_i,j$$

and

$$sum _i B_i,iequiv operatornameTr[B]$$

So, the Mathematica equivalent of:

$$sum _i sum _j sum _m sum _n B_i,j B_j,m B_m,n B_n,i $$

is:

Tr[B . B . B . B]

or:

Tr[MatrixPower[B, 4]]

For the original form of the question, note that:

$$ B_i,jequiv B^T_j,i $$

So, the Mathematica equivalent of:

$$sum _i sum _j sum _k sum _l B_m,i B_m,j B_n,i B_n,j$$

is:

Tr[B.Transpose[B].B.Transpose[B]]

answered 5 hours ago

Carl Woll

58.1k274150

answered 5 hours ago

Carl Woll

58.1k274150

answered 5 hours ago

Carl Woll

58.1k274150

answered 5 hours ago

Carl Woll

58.1k274150

add a commentÂ |Â

up vote
1
down vote

This is a closed matrix formulation of $B^4$.

answered 7 hours ago

David G. Stork

21.5k11646

2

It's the trace of $B^4$, no?
â€“Â LouisB
6 hours ago

add a commentÂ |Â

up vote
1
down vote

This is a closed matrix formulation of $B^4$.

answered 7 hours ago

David G. Stork

21.5k11646

2

It's the trace of $B^4$, no?
â€“Â LouisB
6 hours ago

add a commentÂ |Â

up vote
1
down vote

This is a closed matrix formulation of $B^4$.

answered 7 hours ago

David G. Stork

21.5k11646

This is a closed matrix formulation of $B^4$.

answered 7 hours ago

David G. Stork

21.5k11646

answered 7 hours ago

David G. Stork

21.5k11646

answered 7 hours ago

David G. Stork

21.5k11646

answered 7 hours ago

David G. Stork

21.5k11646

2

It's the trace of $B^4$, no?
â€“Â LouisB
6 hours ago

add a commentÂ |Â

2

It's the trace of $B^4$, no?
â€“Â LouisB
6 hours ago

It's the trace of $B^4$, no?
â€“Â LouisB
6 hours ago

add a commentÂ |Â

up vote
0
down vote

n = 5000;
B = RandomReal[-1, 1, n, n];
a = Tr[B.B.B.B]; // RepeatedTiming // First
b = Tr[MatrixPower[B, 4]]; // RepeatedTiming // First
c = With[A = B.B, Total[A Transpose[A], 2]]; // RepeatedTiming // First
a == b == c

5.7

4.207

2.5

True

For the original question, LouisB's answer (c below) seems to be both correct and efficient:

n = 5000;
B = RandomReal[-1, 1, n, n];
a = Tr[B.Transpose[B].B.Transpose[B]]; // RepeatedTiming // First
b = With[u = Flatten[B.Transpose[B]], u.u]; // RepeatedTiming // First
c = Total[(Transpose[B].B)^2, 2]; // RepeatedTiming // First
a == b == c

5.8

2.4

2.40

True

edited 21 mins ago

answered 34 mins ago

Henrik Schumacher

38.8k254114

add a commentÂ |Â

up vote
0
down vote

n = 5000;
B = RandomReal[-1, 1, n, n];
a = Tr[B.B.B.B]; // RepeatedTiming // First
b = Tr[MatrixPower[B, 4]]; // RepeatedTiming // First
c = With[A = B.B, Total[A Transpose[A], 2]]; // RepeatedTiming // First
a == b == c

5.7

4.207

2.5

True

For the original question, LouisB's answer (c below) seems to be both correct and efficient:

n = 5000;
B = RandomReal[-1, 1, n, n];
a = Tr[B.Transpose[B].B.Transpose[B]]; // RepeatedTiming // First
b = With[u = Flatten[B.Transpose[B]], u.u]; // RepeatedTiming // First
c = Total[(Transpose[B].B)^2, 2]; // RepeatedTiming // First
a == b == c

5.8

2.4

2.40

True

edited 21 mins ago

answered 34 mins ago

Henrik Schumacher

38.8k254114

add a commentÂ |Â

up vote
0
down vote

n = 5000;
B = RandomReal[-1, 1, n, n];
a = Tr[B.B.B.B]; // RepeatedTiming // First
b = Tr[MatrixPower[B, 4]]; // RepeatedTiming // First
c = With[A = B.B, Total[A Transpose[A], 2]]; // RepeatedTiming // First
a == b == c

5.7

4.207

2.5

True

For the original question, LouisB's answer (c below) seems to be both correct and efficient:

n = 5000;
B = RandomReal[-1, 1, n, n];
a = Tr[B.Transpose[B].B.Transpose[B]]; // RepeatedTiming // First
b = With[u = Flatten[B.Transpose[B]], u.u]; // RepeatedTiming // First
c = Total[(Transpose[B].B)^2, 2]; // RepeatedTiming // First
a == b == c

5.8

2.4

2.40

True

edited 21 mins ago

answered 34 mins ago

Henrik Schumacher

38.8k254114

n = 5000;
B = RandomReal[-1, 1, n, n];
a = Tr[B.B.B.B]; // RepeatedTiming // First
b = Tr[MatrixPower[B, 4]]; // RepeatedTiming // First
c = With[A = B.B, Total[A Transpose[A], 2]]; // RepeatedTiming // First
a == b == c

5.7

4.207

2.5

True

For the original question, LouisB's answer (c below) seems to be both correct and efficient:

n = 5000;
B = RandomReal[-1, 1, n, n];
a = Tr[B.Transpose[B].B.Transpose[B]]; // RepeatedTiming // First
b = With[u = Flatten[B.Transpose[B]], u.u]; // RepeatedTiming // First
c = Total[(Transpose[B].B)^2, 2]; // RepeatedTiming // First
a == b == c

5.8

2.4

2.40

True

edited 21 mins ago

answered 34 mins ago

Henrik Schumacher

38.8k254114

edited 21 mins ago

answered 34 mins ago

Henrik Schumacher

38.8k254114

answered 34 mins ago

Henrik Schumacher

38.8k254114

answered 34 mins ago

Henrik Schumacher

38.8k254114

add a commentÂ |Â

farhadpti is a new contributor. Be nice, and check out our Code of Conduct.

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