Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In Shilov's Linear Algebra p.22 about Laplace's theorem

it said

"Finally, let the rows of the determinant $D$ with indices $i_1,i_2,\ldots,i_k$ be fixed; some elements from these rows appear in every term of D."

Why the sentence after ; is true?

In the text, Shilov formed a minor $M^{i_1,i_2,\ldots,i_k}_{j_1,j_2,\ldots,j_k}$ with those k rows and k of the n columns

and a cofactor $\overline{A}^{i_1,i_2,\ldots,i_k}_{j_1,j_2,\ldots.j_k}$ of the minor.

And the terms are now divided into groups $M^{i_1,i_2,\ldots,i_k}_{j_1,j_2,\ldots,j_k} \overline{A}^{i_1,i_2,\ldots,i_k}_{j_1,j_2,\ldots.j_k}$

Note that,

$a_{\alpha_1,1} a_{\alpha_2,2} \cdots a_{\alpha_n,n}$ is a term of $D$, $a_{\alpha_1,1}$ is an element on the first column of the matrix of $D$, $\alpha_1,\alpha_2,\ldots\alpha_n$ are unique.

$a_{i,j}$ is an element of the matrix of $D$.

share|improve this question
    
By the way, your question on the heading has the answer: no, but that question isn't the same as in the body of your question... –  DonAntonio Aug 21 '12 at 5:54
add comment

3 Answers

In fact his follows from the definition of determinant of a square matrix $\,n\times n\,$, which is the sum of $\,n!\,$ products, each one containing exactly one unique element from each row and one unique element from each column of the matrix...

share|improve this answer
    
I think the answer is not even close. –  RHS Aug 21 '12 at 5:58
    
Why don't you think so? That is exactly the answer. –  Robert Israel Aug 21 '12 at 6:03
    
I want to know what is Shilov really talking about. Not defination. –  RHS Aug 21 '12 at 6:09
    
You want to know what Shilov's talking about? I think my answer addresses precisely this, but I understand both the translation, which I think is not the best, and Shilov's notation, which is really frightening, add to the confussion. –  DonAntonio Aug 21 '12 at 11:32
    
Finally, take what the text says and minimize it to one row (or column, by the way): "let the row of the determinant D with index $\,i_1\,$ be fixed: some element of this row appears in every term of D", which seems to be a sloppy way of saying: for any given fixed row of D, every product in the sum in the definition of D contains exactly one, and only one, element of that row as a factor". –  DonAntonio Aug 21 '12 at 11:36
show 3 more comments

Each term of the determinant in fact contains one entry from each row and one entry from each column of the matrix. Think about the way you compute $2\times2$ or $3\times3$ determinants, or look at the general formulas.

share|improve this answer
    
Could you give an example? –  RHS Aug 21 '12 at 5:59
    
In the $3 \times 3$ case, the determinant is $a_{{1,1}}a_{{2,2}}a_{{3,3}}-a_{{1,1}}a_{{2,3}}a_{{3,2}}+a_{{2,1}}a_{{3 ,2}}a_{{1,3}}-a_{{2,1}}a_{{1,2}}a_{{3,3}}+a_{{3,1}}a_{{1,2}}a_{{2,3}}- a_{{3,1}}a_{{2,2}}a_{{1,3}} $. Each term, e.g. $-a_{{2,1}}a_{{1,2}}a_{{3,3}}$, has one entry from each row and one entry from each column. –  Robert Israel Aug 21 '12 at 6:05
    
But you didn't ans why Shilov said every terms. –  RHS Aug 21 '12 at 6:11
    
In Robert's example, there are 6 terms in the determinant. An element from row 1 appears in every one of those 6 terms; an element from row 2 appears in every one of those 6 terms; an element from row 3 appears in every one of those 6 terms. That's what Shilov is saying. –  Gerry Myerson Aug 21 '12 at 6:41
add comment

Consider the minor (part of a term) mentioned above.

The k rows are fixed and k columns $j_1,j_2,\ldots,j_k$ are chosen from n.

So, e.g. in the first row of the minor, would be consist of some element $a_{i_1,j_1},a_{i_11,j_2},\ldots,a_{i_1,j_k}$ from the matrix of $D$.

It is true for every other rows of the minor and other minors.

Because $D$ is consist of the sum of the product of minors and its corresponding cofactor.

So, every term of $D$ is made of of some elements of those k rows.

I think this is what Shilvo trying to said.

share|improve this answer
    
Why don't you think Shilov is trying to say what DonAntonio and Robert Israel and I agree Shilov is trying to say? –  Gerry Myerson Aug 21 '12 at 12:17
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.