Mixing
Non-permutational Definition of the Determinant
$begingroup$
I am looking for a non-permutational definition of determinant. The definition should have these properties:
1: Calculational power (easily applicable, it cold be used for practical calculations).
2: It should hold the usual properties (otherwise it wouldn't be a determinant, right!)
3: No permutation, No permutation, please no permutations.
I would also appreciate if you apply the definition and calculate the determinant of a $4×4$ matrix.
For all I care the definition could be from an obsolete parchment, but it needs to have those three properties.
Appreciate all the help!
linear-algebra definition determinant
asked Jan 22 at 21:28
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483115
$endgroup$
|
show 6 more comments
$begingroup$
I am looking for a non-permutational definition of determinant. The definition should have these properties:
1: Calculational power (easily applicable, it cold be used for practical calculations).
2: It should hold the usual properties (otherwise it wouldn't be a determinant, right!)
3: No permutation, No permutation, please no permutations.
I would also appreciate if you apply the definition and calculate the determinant of a $4×4$ matrix.
For all I care the definition could be from an obsolete parchment, but it needs to have those three properties.
Appreciate all the help!
linear-algebra definition determinant
asked Jan 22 at 21:28
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483115
$endgroup$
1
$begingroup$
Alternating scalar valued tensor with value 1 at the point $(e_1,...,e_n)$. It is equivalent to the permutation definition, so there is no escaping it.
$endgroup$
– copper.hat
Jan 22 at 21:38
1
$begingroup$
You cannot escape the permutations.
$endgroup$
– copper.hat
Jan 22 at 21:38
1
$begingroup$
Then take the Leibniz formula as in my answer. The permutations are not directly visible, but still in the background, like copper.hat says.
$endgroup$
– Dietrich Burde
Jan 22 at 21:39
1
$begingroup$
The determinant of a triangular matrix is the product of the diagonal coefficients. For the general matrix, perform Gauss-Jordan reduction ( en.m.wikipedia.org/wiki/Gaussian_elimination ) till you get a triangular matrix, the determinant of each step being easy to compute. You may also want to look at math.stackexchange.com/questions/668/… .
$endgroup$
– Mindlack
Jan 22 at 21:45
2
$begingroup$
@BertrandWittgenstein'sGhost: At some level, permutations are the essence of the determinant... Being multlinear means there are potentially $2^n$ computations, by being alternating it reduces to $n!$. A smaller number means you are dealing with a special case (Hessenberg, triangular, etc.)
$endgroup$
– copper.hat
Jan 22 at 22:04
|
show 6 more comments
$begingroup$
I am looking for a non-permutational definition of determinant. The definition should have these properties:
1: Calculational power (easily applicable, it cold be used for practical calculations).
2: It should hold the usual properties (otherwise it wouldn't be a determinant, right!)
3: No permutation, No permutation, please no permutations.
I would also appreciate if you apply the definition and calculate the determinant of a $4×4$ matrix.
For all I care the definition could be from an obsolete parchment, but it needs to have those three properties.
Appreciate all the help!
linear-algebra definition determinant
asked Jan 22 at 21:28
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483115
$endgroup$
I am looking for a non-permutational definition of determinant. The definition should have these properties:
1: Calculational power (easily applicable, it cold be used for practical calculations).
2: It should hold the usual properties (otherwise it wouldn't be a determinant, right!)
3: No permutation, No permutation, please no permutations.
I would also appreciate if you apply the definition and calculate the determinant of a $4×4$ matrix.
For all I care the definition could be from an obsolete parchment, but it needs to have those three properties.
Appreciate all the help!
linear-algebra definition determinant
linear-algebra definition determinant
asked Jan 22 at 21:28
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483115
asked Jan 22 at 21:28
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483115
asked Jan 22 at 21:28
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483115
asked Jan 22 at 21:28
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483115
asked Jan 22 at 21:28
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483115
483115
1
$begingroup$
Alternating scalar valued tensor with value 1 at the point $(e_1,...,e_n)$. It is equivalent to the permutation definition, so there is no escaping it.
$endgroup$
– copper.hat
Jan 22 at 21:38
1
$begingroup$
You cannot escape the permutations.
$endgroup$
– copper.hat
Jan 22 at 21:38
1
$begingroup$
Then take the Leibniz formula as in my answer. The permutations are not directly visible, but still in the background, like copper.hat says.
$endgroup$
– Dietrich Burde
Jan 22 at 21:39
1
$begingroup$
The determinant of a triangular matrix is the product of the diagonal coefficients. For the general matrix, perform Gauss-Jordan reduction ( en.m.wikipedia.org/wiki/Gaussian_elimination ) till you get a triangular matrix, the determinant of each step being easy to compute. You may also want to look at math.stackexchange.com/questions/668/… .
$endgroup$
– Mindlack
Jan 22 at 21:45
2
$begingroup$
@BertrandWittgenstein'sGhost: At some level, permutations are the essence of the determinant... Being multlinear means there are potentially $2^n$ computations, by being alternating it reduces to $n!$. A smaller number means you are dealing with a special case (Hessenberg, triangular, etc.)
$endgroup$
– copper.hat
Jan 22 at 22:04
|
show 6 more comments
1
$begingroup$
Alternating scalar valued tensor with value 1 at the point $(e_1,...,e_n)$. It is equivalent to the permutation definition, so there is no escaping it.
$endgroup$
– copper.hat
Jan 22 at 21:38
1
$begingroup$
You cannot escape the permutations.
$endgroup$
– copper.hat
Jan 22 at 21:38
1
$begingroup$
Then take the Leibniz formula as in my answer. The permutations are not directly visible, but still in the background, like copper.hat says.
$endgroup$
– Dietrich Burde
Jan 22 at 21:39
1
$begingroup$
The determinant of a triangular matrix is the product of the diagonal coefficients. For the general matrix, perform Gauss-Jordan reduction ( en.m.wikipedia.org/wiki/Gaussian_elimination ) till you get a triangular matrix, the determinant of each step being easy to compute. You may also want to look at math.stackexchange.com/questions/668/… .
$endgroup$
– Mindlack
Jan 22 at 21:45
2
$begingroup$
@BertrandWittgenstein'sGhost: At some level, permutations are the essence of the determinant... Being multlinear means there are potentially $2^n$ computations, by being alternating it reduces to $n!$. A smaller number means you are dealing with a special case (Hessenberg, triangular, etc.)
$endgroup$
– copper.hat
Jan 22 at 22:04
1
1
$begingroup$
Alternating scalar valued tensor with value 1 at the point $(e_1,...,e_n)$. It is equivalent to the permutation definition, so there is no escaping it.
$endgroup$
– copper.hat
Jan 22 at 21:38
$begingroup$
Alternating scalar valued tensor with value 1 at the point $(e_1,...,e_n)$. It is equivalent to the permutation definition, so there is no escaping it.
$endgroup$
– copper.hat
Jan 22 at 21:38
1
1
$begingroup$
You cannot escape the permutations.
$endgroup$
– copper.hat
Jan 22 at 21:38
$begingroup$
You cannot escape the permutations.
$endgroup$
– copper.hat
Jan 22 at 21:38
1
1
$begingroup$
Then take the Leibniz formula as in my answer. The permutations are not directly visible, but still in the background, like copper.hat says.
$endgroup$
– Dietrich Burde
Jan 22 at 21:39
$begingroup$
Then take the Leibniz formula as in my answer. The permutations are not directly visible, but still in the background, like copper.hat says.
$endgroup$
– Dietrich Burde
Jan 22 at 21:39
1
1
$begingroup$
The determinant of a triangular matrix is the product of the diagonal coefficients. For the general matrix, perform Gauss-Jordan reduction ( en.m.wikipedia.org/wiki/Gaussian_elimination ) till you get a triangular matrix, the determinant of each step being easy to compute. You may also want to look at math.stackexchange.com/questions/668/… .
$endgroup$
– Mindlack
Jan 22 at 21:45
$begingroup$
The determinant of a triangular matrix is the product of the diagonal coefficients. For the general matrix, perform Gauss-Jordan reduction ( en.m.wikipedia.org/wiki/Gaussian_elimination ) till you get a triangular matrix, the determinant of each step being easy to compute. You may also want to look at math.stackexchange.com/questions/668/… .
$endgroup$
– Mindlack
Jan 22 at 21:45
2
2
$begingroup$
@BertrandWittgenstein'sGhost: At some level, permutations are the essence of the determinant... Being multlinear means there are potentially $2^n$ computations, by being alternating it reduces to $n!$. A smaller number means you are dealing with a special case (Hessenberg, triangular, etc.)
$endgroup$
– copper.hat
Jan 22 at 22:04
$begingroup$
@BertrandWittgenstein'sGhost: At some level, permutations are the essence of the determinant... Being multlinear means there are potentially $2^n$ computations, by being alternating it reduces to $n!$. A smaller number means you are dealing with a special case (Hessenberg, triangular, etc.)
$endgroup$
– copper.hat
Jan 22 at 22:04
|
show 6 more comments
1 Answer
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oldest
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$begingroup$
For the $3times 3$-determinant we can use the Rule of Sarrus. Then
the $4times 4$ determinant reduces to the $3times 3$ determinant, because of
$${begin{vmatrix}a&b&c&d\e&f&g&h\i&j&k&l\m&n&o&pend{vmatrix}}=a,{begin{vmatrix}f&g&h\j&k&l\n&o&pend{vmatrix}}-b,{begin{vmatrix}e&g&h\i&k&l\m&o&pend{vmatrix}}+c,{begin{vmatrix}e&f&h\i&j&l\m&n&pend{vmatrix}}-d,{begin{vmatrix}e&f&g\i&j&k\m&n&oend{vmatrix}}.$$
answered Jan 22 at 21:38
Dietrich BurdeDietrich Burde
80.4k647104
$endgroup$
$begingroup$
Thanks for that, but it is not general. I need a general definition. The $4×4$ calculation was just so I can see how the def. Works when applied.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 22 at 21:40
$begingroup$
The general definition is, as already said, by the Weiserstrass axioms. From there $1times 1$, $2times 2$, $3times 3$ are obvious. Then we also have $4times 4$ from the above formula.
$endgroup$
– Dietrich Burde
Jan 23 at 9:05
add a comment |
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$begingroup$
For the $3times 3$-determinant we can use the Rule of Sarrus. Then
the $4times 4$ determinant reduces to the $3times 3$ determinant, because of
$${begin{vmatrix}a&b&c&d\e&f&g&h\i&j&k&l\m&n&o&pend{vmatrix}}=a,{begin{vmatrix}f&g&h\j&k&l\n&o&pend{vmatrix}}-b,{begin{vmatrix}e&g&h\i&k&l\m&o&pend{vmatrix}}+c,{begin{vmatrix}e&f&h\i&j&l\m&n&pend{vmatrix}}-d,{begin{vmatrix}e&f&g\i&j&k\m&n&oend{vmatrix}}.$$
answered Jan 22 at 21:38
Dietrich BurdeDietrich Burde
80.4k647104
$endgroup$
$begingroup$
Thanks for that, but it is not general. I need a general definition. The $4×4$ calculation was just so I can see how the def. Works when applied.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 22 at 21:40
$begingroup$
The general definition is, as already said, by the Weiserstrass axioms. From there $1times 1$, $2times 2$, $3times 3$ are obvious. Then we also have $4times 4$ from the above formula.
$endgroup$
– Dietrich Burde
Jan 23 at 9:05
add a comment |
$begingroup$
For the $3times 3$-determinant we can use the Rule of Sarrus. Then
the $4times 4$ determinant reduces to the $3times 3$ determinant, because of
$${begin{vmatrix}a&b&c&d\e&f&g&h\i&j&k&l\m&n&o&pend{vmatrix}}=a,{begin{vmatrix}f&g&h\j&k&l\n&o&pend{vmatrix}}-b,{begin{vmatrix}e&g&h\i&k&l\m&o&pend{vmatrix}}+c,{begin{vmatrix}e&f&h\i&j&l\m&n&pend{vmatrix}}-d,{begin{vmatrix}e&f&g\i&j&k\m&n&oend{vmatrix}}.$$
answered Jan 22 at 21:38
Dietrich BurdeDietrich Burde
80.4k647104
$endgroup$
$begingroup$
Thanks for that, but it is not general. I need a general definition. The $4×4$ calculation was just so I can see how the def. Works when applied.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 22 at 21:40
$begingroup$
The general definition is, as already said, by the Weiserstrass axioms. From there $1times 1$, $2times 2$, $3times 3$ are obvious. Then we also have $4times 4$ from the above formula.
$endgroup$
– Dietrich Burde
Jan 23 at 9:05
add a comment |
$begingroup$
For the $3times 3$-determinant we can use the Rule of Sarrus. Then
the $4times 4$ determinant reduces to the $3times 3$ determinant, because of
$${begin{vmatrix}a&b&c&d\e&f&g&h\i&j&k&l\m&n&o&pend{vmatrix}}=a,{begin{vmatrix}f&g&h\j&k&l\n&o&pend{vmatrix}}-b,{begin{vmatrix}e&g&h\i&k&l\m&o&pend{vmatrix}}+c,{begin{vmatrix}e&f&h\i&j&l\m&n&pend{vmatrix}}-d,{begin{vmatrix}e&f&g\i&j&k\m&n&oend{vmatrix}}.$$
answered Jan 22 at 21:38
Dietrich BurdeDietrich Burde
80.4k647104
$endgroup$
For the $3times 3$-determinant we can use the Rule of Sarrus. Then
the $4times 4$ determinant reduces to the $3times 3$ determinant, because of
$${begin{vmatrix}a&b&c&d\e&f&g&h\i&j&k&l\m&n&o&pend{vmatrix}}=a,{begin{vmatrix}f&g&h\j&k&l\n&o&pend{vmatrix}}-b,{begin{vmatrix}e&g&h\i&k&l\m&o&pend{vmatrix}}+c,{begin{vmatrix}e&f&h\i&j&l\m&n&pend{vmatrix}}-d,{begin{vmatrix}e&f&g\i&j&k\m&n&oend{vmatrix}}.$$
answered Jan 22 at 21:38
Dietrich BurdeDietrich Burde
80.4k647104
answered Jan 22 at 21:38
Dietrich BurdeDietrich Burde
80.4k647104
answered Jan 22 at 21:38
Dietrich BurdeDietrich Burde
80.4k647104
answered Jan 22 at 21:38
Dietrich BurdeDietrich Burde
80.4k647104
80.4k647104
$begingroup$
Thanks for that, but it is not general. I need a general definition. The $4×4$ calculation was just so I can see how the def. Works when applied.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 22 at 21:40
$begingroup$
The general definition is, as already said, by the Weiserstrass axioms. From there $1times 1$, $2times 2$, $3times 3$ are obvious. Then we also have $4times 4$ from the above formula.
$endgroup$
– Dietrich Burde
Jan 23 at 9:05
add a comment |
$begingroup$
Thanks for that, but it is not general. I need a general definition. The $4×4$ calculation was just so I can see how the def. Works when applied.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 22 at 21:40
$begingroup$
The general definition is, as already said, by the Weiserstrass axioms. From there $1times 1$, $2times 2$, $3times 3$ are obvious. Then we also have $4times 4$ from the above formula.
$endgroup$
– Dietrich Burde
Jan 23 at 9:05
$begingroup$
Thanks for that, but it is not general. I need a general definition. The $4×4$ calculation was just so I can see how the def. Works when applied.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 22 at 21:40
$begingroup$
Thanks for that, but it is not general. I need a general definition. The $4×4$ calculation was just so I can see how the def. Works when applied.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 22 at 21:40
$begingroup$
The general definition is, as already said, by the Weiserstrass axioms. From there $1times 1$, $2times 2$, $3times 3$ are obvious. Then we also have $4times 4$ from the above formula.
$endgroup$
– Dietrich Burde
Jan 23 at 9:05
$begingroup$
The general definition is, as already said, by the Weiserstrass axioms. From there $1times 1$, $2times 2$, $3times 3$ are obvious. Then we also have $4times 4$ from the above formula.
$endgroup$
– Dietrich Burde
Jan 23 at 9:05
add a comment |
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1
$begingroup$
Alternating scalar valued tensor with value 1 at the point $(e_1,...,e_n)$. It is equivalent to the permutation definition, so there is no escaping it.
$endgroup$
– copper.hat
Jan 22 at 21:38
1
$begingroup$
You cannot escape the permutations.
$endgroup$
– copper.hat
Jan 22 at 21:38
1
$begingroup$
Then take the Leibniz formula as in my answer. The permutations are not directly visible, but still in the background, like copper.hat says.
$endgroup$
– Dietrich Burde
Jan 22 at 21:39
1
$begingroup$
The determinant of a triangular matrix is the product of the diagonal coefficients. For the general matrix, perform Gauss-Jordan reduction ( en.m.wikipedia.org/wiki/Gaussian_elimination ) till you get a triangular matrix, the determinant of each step being easy to compute. You may also want to look at math.stackexchange.com/questions/668/… .
$endgroup$
– Mindlack
Jan 22 at 21:45
2
$begingroup$
@BertrandWittgenstein'sGhost: At some level, permutations are the essence of the determinant... Being multlinear means there are potentially $2^n$ computations, by being alternating it reduces to $n!$. A smaller number means you are dealing with a special case (Hessenberg, triangular, etc.)
$endgroup$
– copper.hat
Jan 22 at 22:04