Should the determinant of a $0times 0$ matrix be defined as $0$ or $1$? [duplicate]
This question already has an answer here:
What is the determinant of ? [closed]
2 answers
Is the 0x0 matrix (zero-times-zero matrix) a well-defined concept?
1 answer
Let $A$ be the $0times 0$ matrix and let $det(A)$ be its determinant. I am wondering if $det(A)$ should be defined as $0$ or $1$.
- If we use the definition that determinant of an $ntimes n$ matrix $(a_{ij})$ be defined as $displaystylesum(-1)^{tau(j_1cdots j_n)}a_{1j_1}cdots a_{n{j_n}}$, where $tau(j_1cdots j_n)$ is the inversion number of the permutation $j_1cdots j_n$, then since there is no term present, $det(A)$ should be defined as $0$.
- If we expand the $1times 1$ matrix $(1)$ along the first row, we obtain
$$1=det(1)=1cdotdet(A),$$ which implies that $det(A)$ should be defined as $1$.
Which definition of the determinant of the $0times 0$ matrix $A$, if any, makes more sense here?
definition determinant
asked Jan 4 at 14:41
ZurielZuriel
1,5831028
marked as duplicate by Peter, Dietrich Burde, Ross Millikan, egreg, ccorn Jan 4 at 16:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
|
show 5 more comments
This question already has an answer here:
What is the determinant of ? [closed]
2 answers
Is the 0x0 matrix (zero-times-zero matrix) a well-defined concept?
1 answer
Let $A$ be the $0times 0$ matrix and let $det(A)$ be its determinant. I am wondering if $det(A)$ should be defined as $0$ or $1$.
- If we use the definition that determinant of an $ntimes n$ matrix $(a_{ij})$ be defined as $displaystylesum(-1)^{tau(j_1cdots j_n)}a_{1j_1}cdots a_{n{j_n}}$, where $tau(j_1cdots j_n)$ is the inversion number of the permutation $j_1cdots j_n$, then since there is no term present, $det(A)$ should be defined as $0$.
- If we expand the $1times 1$ matrix $(1)$ along the first row, we obtain
$$1=det(1)=1cdotdet(A),$$ which implies that $det(A)$ should be defined as $1$.
Which definition of the determinant of the $0times 0$ matrix $A$, if any, makes more sense here?
definition determinant
asked Jan 4 at 14:41
ZurielZuriel
1,5831028
marked as duplicate by Peter, Dietrich Burde, Ross Millikan, egreg, ccorn Jan 4 at 16:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
Honest nonsnarky question: why does it matter?
– Randall
Jan 4 at 14:43
The definition of a determinant would give an empty sum, which conventionally is $0$.
– Oscar Lanzi
Jan 4 at 14:44
2
This is very similar to the question as to whether $0^0$ should be zero or one.
– amsmath
Jan 4 at 14:59
2
@Peter Because when we solve this problem for a $0times0$ matrix, we can move to $-1times-1$ matrix :)
– Oldboy
Jan 4 at 16:31
1
@OscarLanzi: The element is an empty product, and thus equals $1$ by definition. The sum is a sum over permutations, not over elements of $left{1,2,ldots,nright}$.
– darij grinberg
Jan 5 at 14:17
|
show 5 more comments
This question already has an answer here:
What is the determinant of ? [closed]
2 answers
Is the 0x0 matrix (zero-times-zero matrix) a well-defined concept?
1 answer
Let $A$ be the $0times 0$ matrix and let $det(A)$ be its determinant. I am wondering if $det(A)$ should be defined as $0$ or $1$.
- If we use the definition that determinant of an $ntimes n$ matrix $(a_{ij})$ be defined as $displaystylesum(-1)^{tau(j_1cdots j_n)}a_{1j_1}cdots a_{n{j_n}}$, where $tau(j_1cdots j_n)$ is the inversion number of the permutation $j_1cdots j_n$, then since there is no term present, $det(A)$ should be defined as $0$.
- If we expand the $1times 1$ matrix $(1)$ along the first row, we obtain
$$1=det(1)=1cdotdet(A),$$ which implies that $det(A)$ should be defined as $1$.
Which definition of the determinant of the $0times 0$ matrix $A$, if any, makes more sense here?
definition determinant
asked Jan 4 at 14:41
ZurielZuriel
1,5831028
This question already has an answer here:
What is the determinant of ? [closed]
2 answers
Is the 0x0 matrix (zero-times-zero matrix) a well-defined concept?
1 answer
Let $A$ be the $0times 0$ matrix and let $det(A)$ be its determinant. I am wondering if $det(A)$ should be defined as $0$ or $1$.
- If we use the definition that determinant of an $ntimes n$ matrix $(a_{ij})$ be defined as $displaystylesum(-1)^{tau(j_1cdots j_n)}a_{1j_1}cdots a_{n{j_n}}$, where $tau(j_1cdots j_n)$ is the inversion number of the permutation $j_1cdots j_n$, then since there is no term present, $det(A)$ should be defined as $0$.
- If we expand the $1times 1$ matrix $(1)$ along the first row, we obtain
$$1=det(1)=1cdotdet(A),$$ which implies that $det(A)$ should be defined as $1$.
Which definition of the determinant of the $0times 0$ matrix $A$, if any, makes more sense here?
This question already has an answer here:
What is the determinant of ? [closed]
2 answers
Is the 0x0 matrix (zero-times-zero matrix) a well-defined concept?
1 answer
definition determinant
definition determinant
asked Jan 4 at 14:41
ZurielZuriel
1,5831028
asked Jan 4 at 14:41
ZurielZuriel
1,5831028
asked Jan 4 at 14:41
ZurielZuriel
1,5831028
asked Jan 4 at 14:41
ZurielZuriel
1,5831028
asked Jan 4 at 14:41
ZurielZuriel
1,5831028
1,5831028
marked as duplicate by Peter, Dietrich Burde, Ross Millikan, egreg, ccorn Jan 4 at 16:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Peter, Dietrich Burde, Ross Millikan, egreg, ccorn Jan 4 at 16:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
Honest nonsnarky question: why does it matter?
– Randall
Jan 4 at 14:43
The definition of a determinant would give an empty sum, which conventionally is $0$.
– Oscar Lanzi
Jan 4 at 14:44
2
This is very similar to the question as to whether $0^0$ should be zero or one.
– amsmath
Jan 4 at 14:59
2
@Peter Because when we solve this problem for a $0times0$ matrix, we can move to $-1times-1$ matrix :)
– Oldboy
Jan 4 at 16:31
1
@OscarLanzi: The element is an empty product, and thus equals $1$ by definition. The sum is a sum over permutations, not over elements of $left{1,2,ldots,nright}$.
– darij grinberg
Jan 5 at 14:17
|
show 5 more comments
3
Honest nonsnarky question: why does it matter?
– Randall
Jan 4 at 14:43
The definition of a determinant would give an empty sum, which conventionally is $0$.
– Oscar Lanzi
Jan 4 at 14:44
2
This is very similar to the question as to whether $0^0$ should be zero or one.
– amsmath
Jan 4 at 14:59
2
@Peter Because when we solve this problem for a $0times0$ matrix, we can move to $-1times-1$ matrix :)
– Oldboy
Jan 4 at 16:31
1
@OscarLanzi: The element is an empty product, and thus equals $1$ by definition. The sum is a sum over permutations, not over elements of $left{1,2,ldots,nright}$.
– darij grinberg
Jan 5 at 14:17
3
3
Honest nonsnarky question: why does it matter?
– Randall
Jan 4 at 14:43
Honest nonsnarky question: why does it matter?
– Randall
Jan 4 at 14:43
The definition of a determinant would give an empty sum, which conventionally is $0$.
– Oscar Lanzi
Jan 4 at 14:44
The definition of a determinant would give an empty sum, which conventionally is $0$.
– Oscar Lanzi
Jan 4 at 14:44
2
2
This is very similar to the question as to whether $0^0$ should be zero or one.
– amsmath
Jan 4 at 14:59
This is very similar to the question as to whether $0^0$ should be zero or one.
– amsmath
Jan 4 at 14:59
2
2
@Peter Because when we solve this problem for a $0times0$ matrix, we can move to $-1times-1$ matrix :)
– Oldboy
Jan 4 at 16:31
@Peter Because when we solve this problem for a $0times0$ matrix, we can move to $-1times-1$ matrix :)
– Oldboy
Jan 4 at 16:31
1
1
@OscarLanzi: The element is an empty product, and thus equals $1$ by definition. The sum is a sum over permutations, not over elements of $left{1,2,ldots,nright}$.
– darij grinberg
Jan 5 at 14:17
@OscarLanzi: The element is an empty product, and thus equals $1$ by definition. The sum is a sum over permutations, not over elements of $left{1,2,ldots,nright}$.
– darij grinberg
Jan 5 at 14:17
|
show 5 more comments
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3
Honest nonsnarky question: why does it matter?
– Randall
Jan 4 at 14:43
The definition of a determinant would give an empty sum, which conventionally is $0$.
– Oscar Lanzi
Jan 4 at 14:44
2
This is very similar to the question as to whether $0^0$ should be zero or one.
– amsmath
Jan 4 at 14:59
2
@Peter Because when we solve this problem for a $0times0$ matrix, we can move to $-1times-1$ matrix :)
– Oldboy
Jan 4 at 16:31
1
@OscarLanzi: The element is an empty product, and thus equals $1$ by definition. The sum is a sum over permutations, not over elements of $left{1,2,ldots,nright}$.
– darij grinberg
Jan 5 at 14:17