Generalized Diagonal
I was given the following definition:
For all $sigmain S_n$, the product $prod_limits{i=1}^n a_{i,sigma(i)}$ contains one entry from every row and every column, the entries of these matrix is called a "Generalized Diagonal".
What is Generalized Diagonal? is it all the entries of a given $sigmain S_n$? what are the uses of it?
linear-algebra
asked Jul 22 '15 at 16:20
gbox
5,36662157
add a comment |
I was given the following definition:
For all $sigmain S_n$, the product $prod_limits{i=1}^n a_{i,sigma(i)}$ contains one entry from every row and every column, the entries of these matrix is called a "Generalized Diagonal".
What is Generalized Diagonal? is it all the entries of a given $sigmain S_n$? what are the uses of it?
linear-algebra
asked Jul 22 '15 at 16:20
gbox
5,36662157
add a comment |
I was given the following definition:
For all $sigmain S_n$, the product $prod_limits{i=1}^n a_{i,sigma(i)}$ contains one entry from every row and every column, the entries of these matrix is called a "Generalized Diagonal".
What is Generalized Diagonal? is it all the entries of a given $sigmain S_n$? what are the uses of it?
linear-algebra
asked Jul 22 '15 at 16:20
gbox
5,36662157
I was given the following definition:
For all $sigmain S_n$, the product $prod_limits{i=1}^n a_{i,sigma(i)}$ contains one entry from every row and every column, the entries of these matrix is called a "Generalized Diagonal".
What is Generalized Diagonal? is it all the entries of a given $sigmain S_n$? what are the uses of it?
linear-algebra
linear-algebra
asked Jul 22 '15 at 16:20
gbox
5,36662157
asked Jul 22 '15 at 16:20
gbox
5,36662157
asked Jul 22 '15 at 16:20
gbox
5,36662157
asked Jul 22 '15 at 16:20
gbox
5,36662157
asked Jul 22 '15 at 16:20
gbox
5,36662157
5,36662157
add a comment |
add a comment |
2 Answers
2
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I guess it means the sequence $left(a_{1,sigmaleft(1right)}, a_{2,sigmaleft(2right)}, ldots, a_{n,sigmaleft(nright)}right)$ for a given $sigma in S_n$. I have never heard of this notion; it is not standard. But it isn't completely useless: For instance, it allows you to say that "the determinant of a lower-triangular matrix is the product of its diagonal entries, because every generalized diagonal other than the main diagonal has at least one zero entry".
answered Jul 22 '15 at 17:02
darij grinberg
10.2k33061
add a comment |
This resembles the notation used in the Leibniz formula for determinants.
If I define the signed generalized diagonal product as
$$operatorname{SGDP}_sigma = operatorname{sign}(sigma) prod_limits{i=1}^n a_{i,sigma(i)}$$
where $operatorname{sign}(sigma) = +1$ if $sigma$ is an even permutation and $operatorname{sign}(sigma) = -1$ if $sigma$ is an odd permutation, then I can make the statement:
The determinant of a matrix is equal to the sum of all its signed generalized diagonal products
Or simply, $det(A) = sum_{sigma in S_n} operatorname{SGDP}_sigma$
As a concrete example, here is the formula for the determinant of a $3 times 3$ matrix:
$$det(A) = a_{1,1}a_{2,2}a_{3,3} - a_{1,2}a_{2,1}a_{3,3} - a_{1,1}a_{2,3}a_{3,2} + a_{1,2}a_{2,3}a_{3,1} + a_{1,3}a_{2,1}a_{3,2} - a_{1,3}a_{2,2}a_{3,1} $$
edited Nov 20 at 0:48
darij grinberg
10.2k33061
answered Jul 22 '15 at 17:36
eigenchris
1,520616
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
I guess it means the sequence $left(a_{1,sigmaleft(1right)}, a_{2,sigmaleft(2right)}, ldots, a_{n,sigmaleft(nright)}right)$ for a given $sigma in S_n$. I have never heard of this notion; it is not standard. But it isn't completely useless: For instance, it allows you to say that "the determinant of a lower-triangular matrix is the product of its diagonal entries, because every generalized diagonal other than the main diagonal has at least one zero entry".
answered Jul 22 '15 at 17:02
darij grinberg
10.2k33061
add a comment |
I guess it means the sequence $left(a_{1,sigmaleft(1right)}, a_{2,sigmaleft(2right)}, ldots, a_{n,sigmaleft(nright)}right)$ for a given $sigma in S_n$. I have never heard of this notion; it is not standard. But it isn't completely useless: For instance, it allows you to say that "the determinant of a lower-triangular matrix is the product of its diagonal entries, because every generalized diagonal other than the main diagonal has at least one zero entry".
answered Jul 22 '15 at 17:02
darij grinberg
10.2k33061
add a comment |
I guess it means the sequence $left(a_{1,sigmaleft(1right)}, a_{2,sigmaleft(2right)}, ldots, a_{n,sigmaleft(nright)}right)$ for a given $sigma in S_n$. I have never heard of this notion; it is not standard. But it isn't completely useless: For instance, it allows you to say that "the determinant of a lower-triangular matrix is the product of its diagonal entries, because every generalized diagonal other than the main diagonal has at least one zero entry".
answered Jul 22 '15 at 17:02
darij grinberg
10.2k33061
I guess it means the sequence $left(a_{1,sigmaleft(1right)}, a_{2,sigmaleft(2right)}, ldots, a_{n,sigmaleft(nright)}right)$ for a given $sigma in S_n$. I have never heard of this notion; it is not standard. But it isn't completely useless: For instance, it allows you to say that "the determinant of a lower-triangular matrix is the product of its diagonal entries, because every generalized diagonal other than the main diagonal has at least one zero entry".
answered Jul 22 '15 at 17:02
darij grinberg
10.2k33061
answered Jul 22 '15 at 17:02
darij grinberg
10.2k33061
answered Jul 22 '15 at 17:02
darij grinberg
10.2k33061
answered Jul 22 '15 at 17:02
darij grinberg
10.2k33061
10.2k33061
add a comment |
add a comment |
This resembles the notation used in the Leibniz formula for determinants.
If I define the signed generalized diagonal product as
$$operatorname{SGDP}_sigma = operatorname{sign}(sigma) prod_limits{i=1}^n a_{i,sigma(i)}$$
where $operatorname{sign}(sigma) = +1$ if $sigma$ is an even permutation and $operatorname{sign}(sigma) = -1$ if $sigma$ is an odd permutation, then I can make the statement:
The determinant of a matrix is equal to the sum of all its signed generalized diagonal products
Or simply, $det(A) = sum_{sigma in S_n} operatorname{SGDP}_sigma$
As a concrete example, here is the formula for the determinant of a $3 times 3$ matrix:
$$det(A) = a_{1,1}a_{2,2}a_{3,3} - a_{1,2}a_{2,1}a_{3,3} - a_{1,1}a_{2,3}a_{3,2} + a_{1,2}a_{2,3}a_{3,1} + a_{1,3}a_{2,1}a_{3,2} - a_{1,3}a_{2,2}a_{3,1} $$
edited Nov 20 at 0:48
darij grinberg
10.2k33061
answered Jul 22 '15 at 17:36
eigenchris
1,520616
add a comment |
This resembles the notation used in the Leibniz formula for determinants.
If I define the signed generalized diagonal product as
$$operatorname{SGDP}_sigma = operatorname{sign}(sigma) prod_limits{i=1}^n a_{i,sigma(i)}$$
where $operatorname{sign}(sigma) = +1$ if $sigma$ is an even permutation and $operatorname{sign}(sigma) = -1$ if $sigma$ is an odd permutation, then I can make the statement:
The determinant of a matrix is equal to the sum of all its signed generalized diagonal products
Or simply, $det(A) = sum_{sigma in S_n} operatorname{SGDP}_sigma$
As a concrete example, here is the formula for the determinant of a $3 times 3$ matrix:
$$det(A) = a_{1,1}a_{2,2}a_{3,3} - a_{1,2}a_{2,1}a_{3,3} - a_{1,1}a_{2,3}a_{3,2} + a_{1,2}a_{2,3}a_{3,1} + a_{1,3}a_{2,1}a_{3,2} - a_{1,3}a_{2,2}a_{3,1} $$
edited Nov 20 at 0:48
darij grinberg
10.2k33061
answered Jul 22 '15 at 17:36
eigenchris
1,520616
add a comment |
This resembles the notation used in the Leibniz formula for determinants.
If I define the signed generalized diagonal product as
$$operatorname{SGDP}_sigma = operatorname{sign}(sigma) prod_limits{i=1}^n a_{i,sigma(i)}$$
where $operatorname{sign}(sigma) = +1$ if $sigma$ is an even permutation and $operatorname{sign}(sigma) = -1$ if $sigma$ is an odd permutation, then I can make the statement:
The determinant of a matrix is equal to the sum of all its signed generalized diagonal products
Or simply, $det(A) = sum_{sigma in S_n} operatorname{SGDP}_sigma$
As a concrete example, here is the formula for the determinant of a $3 times 3$ matrix:
$$det(A) = a_{1,1}a_{2,2}a_{3,3} - a_{1,2}a_{2,1}a_{3,3} - a_{1,1}a_{2,3}a_{3,2} + a_{1,2}a_{2,3}a_{3,1} + a_{1,3}a_{2,1}a_{3,2} - a_{1,3}a_{2,2}a_{3,1} $$
edited Nov 20 at 0:48
darij grinberg
10.2k33061
answered Jul 22 '15 at 17:36
eigenchris
1,520616
This resembles the notation used in the Leibniz formula for determinants.
If I define the signed generalized diagonal product as
$$operatorname{SGDP}_sigma = operatorname{sign}(sigma) prod_limits{i=1}^n a_{i,sigma(i)}$$
where $operatorname{sign}(sigma) = +1$ if $sigma$ is an even permutation and $operatorname{sign}(sigma) = -1$ if $sigma$ is an odd permutation, then I can make the statement:
The determinant of a matrix is equal to the sum of all its signed generalized diagonal products
Or simply, $det(A) = sum_{sigma in S_n} operatorname{SGDP}_sigma$
As a concrete example, here is the formula for the determinant of a $3 times 3$ matrix:
$$det(A) = a_{1,1}a_{2,2}a_{3,3} - a_{1,2}a_{2,1}a_{3,3} - a_{1,1}a_{2,3}a_{3,2} + a_{1,2}a_{2,3}a_{3,1} + a_{1,3}a_{2,1}a_{3,2} - a_{1,3}a_{2,2}a_{3,1} $$
edited Nov 20 at 0:48
darij grinberg
10.2k33061
answered Jul 22 '15 at 17:36
eigenchris
1,520616
edited Nov 20 at 0:48
darij grinberg
10.2k33061
edited Nov 20 at 0:48
darij grinberg
10.2k33061
edited Nov 20 at 0:48
darij grinberg
10.2k33061
10.2k33061
answered Jul 22 '15 at 17:36
eigenchris
1,520616
answered Jul 22 '15 at 17:36
eigenchris
1,520616
answered Jul 22 '15 at 17:36
eigenchris
1,520616
1,520616
add a comment |
add a comment |
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