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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 $D$.

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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
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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...

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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
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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.

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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
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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.

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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
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