 |
|
| Science Forum Index » Mathematics Forum » determinant definition...can't figure out what the... |
|
Page 1 of 1 |
|
| Author |
Message |
| genericaudioperson... |
 Posted: Fri Nov 06, 2009 5:02 pm |
|
|
|
Guest
|
Hello,
I am studying a definition of a determinant in a textbook. The left
the index of summation out, which makes it very confusing. I am going
to type the definition they wrote here. I am hoping someone could
tell me what the index should be. Usually there is something like i=1
underneath the big sigma symbol, and then an n above the symbol (which
would tell you to sum the terms one by one up to the number n). But
they did put any index of summation in the definition. Here it is:
"det(A) = Ó (+/-) (a1j1)(a2j2)...(anjn) where the summation is over
all permutations j1j2...jn of the set S={1,2,....n}. The sign is
taken as + or - according to whether the permutation j1j2...jn is even
or odd."
Is the index of summation supposed to be j=1 to n? I can't understand
why the left this information off.
Note: the numbers in a1j1 etc. in the definition above are supposed to
be subscripts, but I don't know how to type those into this type of
newsgroup text). |
|
|
| Back to top |
|
|
|
| Arturo Magidin... |
| Posted: Fri Nov 06, 2009 5:11 pm |
|
|
|
Guest
|
On Nov 6, 9:02 pm, genericaudioperson <genericaudioper... at (no spam) hotmail.com>
wrote:
[quote]Hello,
I am studying a definition of a determinant in a textbook. The left
the index of summation out, which makes it very confusing. I am going
to type the definition they wrote here. I am hoping someone could
tell me what the index should be. Usually there is something like i=1
underneath the big sigma symbol, and then an n above the symbol (which
would tell you to sum the terms one by one up to the number n). But
they did put any index of summation in the definition. Here it is:
"det(A) = Ó (+/-) (a1j1)(a2j2)...(anjn) where the summation is over
all permutations j1j2...jn of the set S={1,2,....n}. The sign is
taken as + or - according to whether the permutation j1j2...jn is even
or odd."
Is the index of summation supposed to be j=1 to n? I can't understand
why the left this information off.
[/quote]
The index is, as is quite clearly stated, "the set of all
permutations". The index is over all j that are bijective functions
from S to itself.
So j is not a number: it's a function. They didn't leave the
information off.
For example, in a 3x3 matrix, you will look at the set S={1,2,3}.
There are six permutaitons (functions from S to S that are bijective);
let's call them I, r, r^2, x, y and z. They are:
I: 1|->1, 2|->2, 3|->3.
r: 1|->2, 2|->3, 3|->1.
r^2 : r|->3, 2|->1, 3|->2.
x: 1|->1, 2|->3, 3|->2.
y: 1|->3, 2|->2, 3|->1.
z: 1|->2, 2|->1, 3|->3.
The set of all permutations is then {I, r, r^2, x, y, z}. The sign of
these permutaions is: x, y, and z are odd permutations, I, r, and r^2
are even permutations.
So the sum is taken over all j in {I, r, r^2, x, y, z}. When j=I, you
get the term a_{11}a_{22}a_{33}. When j=r, you get the term a_{12}a_
{23}a_{31} (the second index is the image of the first under the
permutation you are looking at). When j=r^2, you get a_{13}a_{21}a_
{32}. When j=x you get (-1)a_{11}a_{23}a_{32}. When j=y you get (-1)a_
{13}a_{22}a_{31}. When j=z, you get (-1)a_{12}a_{21}a_{33}.
So the summation, taken over all j in {I, r, r^2, x, y, z}, will
yield:
det(A) = a_{11}a_{22}a_{33} [for j=I]
+ a_{12}a_{23}a_{31} [for j=r]
+ a_{13}a_{21}a_{32} [for j=r^2]
- a_{11}a_{23}a_{32} [for j=x]
- a_{13}a_{22}a_{31} [for j=y]
- a_{12}a_{21}a_{33} [for j=z].
[quote]Note: the numbers in a1j1 etc. in the definition above are supposed to
be subscripts, but I don't know how to type those into this type of
newsgroup text).
[/quote]
The standard way is "pseudo-TeX": _ indicates a subscript, as I did
above.
--
Arturo Magidin |
|
|
| Back to top |
|
|
|
| JEMebius... |
| Posted: Sat Nov 07, 2009 3:30 pm |
|
|
|
Guest
|
genericaudioperson wrote:
[quote]Hello,
I am studying a definition of a determinant in a textbook. The left
the index of summation out, which makes it very confusing. I am going
to type the definition they wrote here. I am hoping someone could
tell me what the index should be. Usually there is something like i=1
underneath the big sigma symbol, and then an n above the symbol (which
would tell you to sum the terms one by one up to the number n). But
they did put any index of summation in the definition. Here it is:
"det(A) = Ó (+/-) (a1j1)(a2j2)...(anjn) where the summation is over
all permutations j1j2...jn of the set S={1,2,....n}. The sign is
taken as + or - according to whether the permutation j1j2...jn is even
or odd."
Is the index of summation supposed to be j=1 to n? I can't understand
why the left this information off.
Note: the numbers in a1j1 etc. in the definition above are supposed to
be subscripts, but I don't know how to type those into this type of
newsgroup text).
[/quote]
Start with observing how things go for orders 2 and 3.
det(A) = A11.A22 - A12.A21
det(A) = A11.A22.A33 + A12.A23.A31 + A13.A21.A32 - A11.A23.A32 - A12.A21.A33 - A13.A22.A31
In the matrix elements Aij there is the second index j.
In the summation there is not a single index j, but rather a multi-index [j1, j2, ... jn].
This multi-index is spread out, so to say, over all factors of each product in the
determinant formula. It consists of the second indexes of the factors.
It takes as its values all permutations of 1...n.
Now turning to the 3rd-order formula:
the multi-index takes the values 1,2,3; 2,3,1; 3,1,2 and 1,3,2; 2,1,3; 3,2,1.
the first three are the even permutations of 1,2,3; the remaining three are the odd
permutations.
Please also read http://en.wikipedia.org/wiki/Determinant - you will find there exactly
the definition of your textbook. The j_i is there represented by sigma(i).
Take note of the rule of Sarrus, which refers specifically to 3rd-order determinants.
Good luck: Johan E. Mebius |
|
|
| Back to top |
|
|
|
| genericaudioperson... |
| Posted: Thu Nov 12, 2009 7:13 pm |
|
|
|
Guest
|
Thanks for the insights.
I guess since you have to some over multiple scenarios, that you can't
simply place an i=1 on the bottom and an n on the top.
Still, I don't like "blind" summations, where the index is not
specified. Having to include an external sentence: "Where the
summation is taken over over all permutations j1j2...jn of the set S=
{1,2,...n}" seems awkward. It also goes against the mathematician's
sensibilities of concision and being able to represent operations
symbolically rather than descriptively.
So maybe someone with some authority in the mathematics world could
create some sort of summation index to place above and below the sigma
which would declare this information, rather than having to create an
elaborate external sentence.
Leibniz used to spend days thinking about a math symbol before he
would commit to promoting one. I could without looking up the history
of mathematics guess with confidence that the determinant was
discovered and promoted after Leibniz passed on. Because he certainly
would have found a way to represent the information in an elegant
symbolic structure.
Sure, there is the Laplace definition, which uses a proper index of
summation and is excellent. But the (a1j1)(a2j2)...(anjn) version
really could use a symbol wizard to get to its proper place
symbolically. |
|
|
| Back to top |
|
|
|
| Robert Israel... |
| Posted: Fri Nov 13, 2009 12:53 am |
|
|
|
Guest
|
genericaudioperson <genericaudioperson at (no spam) hotmail.com> writes:
[quote]Thanks for the insights.
I guess since you have to some over multiple scenarios, that you can't
simply place an i=1 on the bottom and an n on the top.
Still, I don't like "blind" summations, where the index is not
specified. Having to include an external sentence: "Where the
summation is taken over over all permutations j1j2...jn of the set S=
{1,2,...n}" seems awkward. It also goes against the mathematician's
sensibilities of concision and being able to represent operations
symbolically rather than descriptively.
So maybe someone with some authority in the mathematics world could
create some sort of summation index to place above and below the sigma
which would declare this information, rather than having to create an
elaborate external sentence.
[/quote]
Of course:
sum_{s in S_n} sgn(s) product_{i=1}^n a_{i,s(i)}
(S_n being the set of permutations of {1...n} and sgn(s) the signature
of the permutation s).
--
Robert Israel israel at (no spam) math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada |
|
|
| Back to top |
|
|
|
|
|
All times are GMT - 5 Hours
The time now is Mon Dec 07, 2009 7:02 pm
|
|