Mixing
Significance of this note about definition of determinant in Artin's Algebra
$begingroup$
This question is from Artin Algebra, second edition.
After defining determinant and proving many of its properties, author comments about possible alternative approach of defining determinant on P23:
Note: It is natural idea to try defining determinants using compatibility with multiplication and corollary 1.4.13. Since we can write an invertible matrix as a product of elementary matrices, these properties determine the determinants of every invertible matrix. But there are many ways to write a given matrix as such a product. Without going through some steps as we have, it won't be clear that two such products will give the same answer. It isn't easy to make this idea work.
Here, corollary 1.4.13 is this:
But, I can't find the significance of the above note: I have this reasoning: If $A=E_1cdots E_n=E'_1cdots E'_k$, but if $delta(E_1cdots E_n)=delta(E_1)cdots delta(E_n)$ did not equal to $delta(E'_1cdots E'_k)=delta(E'_1)cdots delta(E'_k)$, this would mean that $delta(A)$ has two different values, but $delta$ being function, this is absurd.
(I know that the information I provided for this question is not enough, but things are scattered from P20 to P23 of the book. Hence I urge you to refer Artin's Algebra if you have it. I have been wondering about this note for about a year.)
linear-algebra definition determinant
asked Jan 18 at 14:27
SilentSilent
2,84132152
$endgroup$
add a comment |
$begingroup$
This question is from Artin Algebra, second edition.
After defining determinant and proving many of its properties, author comments about possible alternative approach of defining determinant on P23:
Note: It is natural idea to try defining determinants using compatibility with multiplication and corollary 1.4.13. Since we can write an invertible matrix as a product of elementary matrices, these properties determine the determinants of every invertible matrix. But there are many ways to write a given matrix as such a product. Without going through some steps as we have, it won't be clear that two such products will give the same answer. It isn't easy to make this idea work.
Here, corollary 1.4.13 is this:
But, I can't find the significance of the above note: I have this reasoning: If $A=E_1cdots E_n=E'_1cdots E'_k$, but if $delta(E_1cdots E_n)=delta(E_1)cdots delta(E_n)$ did not equal to $delta(E'_1cdots E'_k)=delta(E'_1)cdots delta(E'_k)$, this would mean that $delta(A)$ has two different values, but $delta$ being function, this is absurd.
(I know that the information I provided for this question is not enough, but things are scattered from P20 to P23 of the book. Hence I urge you to refer Artin's Algebra if you have it. I have been wondering about this note for about a year.)
linear-algebra definition determinant
asked Jan 18 at 14:27
SilentSilent
2,84132152
$endgroup$
$begingroup$
What you think to be absurd is precisely the point: how do you know that $delta$ is well-defined for non-elementary matrices in the first place?
$endgroup$
– user1551
Jan 18 at 14:48
$begingroup$
@user1551, well, corollary 1.4.13 says that $delta$ is a function. Although, I would appreciate if you explain a little bit more.
$endgroup$
– Silent
Jan 18 at 14:50
3
$begingroup$
No, the corollary says that if $delta$ exists and satisfies some properties, then you can make some assertions about it. But it doesn't guarantee the existence of such a $delta$.
$endgroup$
– user1551
Jan 18 at 14:52
$begingroup$
Oh! I think I get your point. Thanks a lot.
$endgroup$
– Silent
Jan 18 at 17:41
add a comment |
$begingroup$
This question is from Artin Algebra, second edition.
After defining determinant and proving many of its properties, author comments about possible alternative approach of defining determinant on P23:
Note: It is natural idea to try defining determinants using compatibility with multiplication and corollary 1.4.13. Since we can write an invertible matrix as a product of elementary matrices, these properties determine the determinants of every invertible matrix. But there are many ways to write a given matrix as such a product. Without going through some steps as we have, it won't be clear that two such products will give the same answer. It isn't easy to make this idea work.
Here, corollary 1.4.13 is this:
But, I can't find the significance of the above note: I have this reasoning: If $A=E_1cdots E_n=E'_1cdots E'_k$, but if $delta(E_1cdots E_n)=delta(E_1)cdots delta(E_n)$ did not equal to $delta(E'_1cdots E'_k)=delta(E'_1)cdots delta(E'_k)$, this would mean that $delta(A)$ has two different values, but $delta$ being function, this is absurd.
(I know that the information I provided for this question is not enough, but things are scattered from P20 to P23 of the book. Hence I urge you to refer Artin's Algebra if you have it. I have been wondering about this note for about a year.)
linear-algebra definition determinant
asked Jan 18 at 14:27
SilentSilent
2,84132152
$endgroup$
This question is from Artin Algebra, second edition.
After defining determinant and proving many of its properties, author comments about possible alternative approach of defining determinant on P23:
Note: It is natural idea to try defining determinants using compatibility with multiplication and corollary 1.4.13. Since we can write an invertible matrix as a product of elementary matrices, these properties determine the determinants of every invertible matrix. But there are many ways to write a given matrix as such a product. Without going through some steps as we have, it won't be clear that two such products will give the same answer. It isn't easy to make this idea work.
Here, corollary 1.4.13 is this:
But, I can't find the significance of the above note: I have this reasoning: If $A=E_1cdots E_n=E'_1cdots E'_k$, but if $delta(E_1cdots E_n)=delta(E_1)cdots delta(E_n)$ did not equal to $delta(E'_1cdots E'_k)=delta(E'_1)cdots delta(E'_k)$, this would mean that $delta(A)$ has two different values, but $delta$ being function, this is absurd.
(I know that the information I provided for this question is not enough, but things are scattered from P20 to P23 of the book. Hence I urge you to refer Artin's Algebra if you have it. I have been wondering about this note for about a year.)
linear-algebra definition determinant
linear-algebra definition determinant
asked Jan 18 at 14:27
SilentSilent
2,84132152
asked Jan 18 at 14:27
SilentSilent
2,84132152
asked Jan 18 at 14:27
SilentSilent
2,84132152
asked Jan 18 at 14:27
SilentSilent
2,84132152
asked Jan 18 at 14:27
SilentSilent
2,84132152
2,84132152
$begingroup$
What you think to be absurd is precisely the point: how do you know that $delta$ is well-defined for non-elementary matrices in the first place?
$endgroup$
– user1551
Jan 18 at 14:48
$begingroup$
@user1551, well, corollary 1.4.13 says that $delta$ is a function. Although, I would appreciate if you explain a little bit more.
$endgroup$
– Silent
Jan 18 at 14:50
3
$begingroup$
No, the corollary says that if $delta$ exists and satisfies some properties, then you can make some assertions about it. But it doesn't guarantee the existence of such a $delta$.
$endgroup$
– user1551
Jan 18 at 14:52
$begingroup$
Oh! I think I get your point. Thanks a lot.
$endgroup$
– Silent
Jan 18 at 17:41
add a comment |
$begingroup$
What you think to be absurd is precisely the point: how do you know that $delta$ is well-defined for non-elementary matrices in the first place?
$endgroup$
– user1551
Jan 18 at 14:48
$begingroup$
@user1551, well, corollary 1.4.13 says that $delta$ is a function. Although, I would appreciate if you explain a little bit more.
$endgroup$
– Silent
Jan 18 at 14:50
3
$begingroup$
No, the corollary says that if $delta$ exists and satisfies some properties, then you can make some assertions about it. But it doesn't guarantee the existence of such a $delta$.
$endgroup$
– user1551
Jan 18 at 14:52
$begingroup$
Oh! I think I get your point. Thanks a lot.
$endgroup$
– Silent
Jan 18 at 17:41
$begingroup$
What you think to be absurd is precisely the point: how do you know that $delta$ is well-defined for non-elementary matrices in the first place?
$endgroup$
– user1551
Jan 18 at 14:48
$begingroup$
What you think to be absurd is precisely the point: how do you know that $delta$ is well-defined for non-elementary matrices in the first place?
$endgroup$
– user1551
Jan 18 at 14:48
$begingroup$
@user1551, well, corollary 1.4.13 says that $delta$ is a function. Although, I would appreciate if you explain a little bit more.
$endgroup$
– Silent
Jan 18 at 14:50
$begingroup$
@user1551, well, corollary 1.4.13 says that $delta$ is a function. Although, I would appreciate if you explain a little bit more.
$endgroup$
– Silent
Jan 18 at 14:50
3
3
$begingroup$
No, the corollary says that if $delta$ exists and satisfies some properties, then you can make some assertions about it. But it doesn't guarantee the existence of such a $delta$.
$endgroup$
– user1551
Jan 18 at 14:52
$begingroup$
No, the corollary says that if $delta$ exists and satisfies some properties, then you can make some assertions about it. But it doesn't guarantee the existence of such a $delta$.
$endgroup$
– user1551
Jan 18 at 14:52
$begingroup$
Oh! I think I get your point. Thanks a lot.
$endgroup$
– Silent
Jan 18 at 17:41
$begingroup$
Oh! I think I get your point. Thanks a lot.
$endgroup$
– Silent
Jan 18 at 17:41
add a comment |
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$begingroup$
What you think to be absurd is precisely the point: how do you know that $delta$ is well-defined for non-elementary matrices in the first place?
$endgroup$
– user1551
Jan 18 at 14:48
$begingroup$
@user1551, well, corollary 1.4.13 says that $delta$ is a function. Although, I would appreciate if you explain a little bit more.
$endgroup$
– Silent
Jan 18 at 14:50
3
$begingroup$
No, the corollary says that if $delta$ exists and satisfies some properties, then you can make some assertions about it. But it doesn't guarantee the existence of such a $delta$.
$endgroup$
– user1551
Jan 18 at 14:52
$begingroup$
Oh! I think I get your point. Thanks a lot.
$endgroup$
– Silent
Jan 18 at 17:41