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So, I am very interested on this theorem (Laplace expansion), but I am still a high school student. I have four books about matrices, but only one of them have proo and that doesn't start with my definition of determinant. Also I don't understand the proof of Laplace expansion in wikipedia, because there is too much symbol and terms I didn't learn and that proof dosn't start with my definition. So may someone give me the outline of the proof start with my definition?

Here's my definition of the $n \times n$ determinant:

The value of the determinant of a matrix of order $n$ is defined as the sum of $n!$ terms of the form $(-1)^k a_{1 i_1} a_{2 i_2} \cdots a_{n i_n}$. Each term contains one and only one element from each row and one and only element from each column; i.e., the second subscripts $i_1, i_2 , \ldots, i_n$ are equal to $1,2, \ldots, n$ taken in some order. The exponent $k$ represents the number of interchanges of two elements necessary for the second subscripts to be placed in the order $1,2, \ldots, n$. For example, consider the term containing $a_{13} a_{21} a_{34} a_{42}$ in the evaluation of the determinant of a matrix of order four. The value of $k$ is $3$ since three interchanges of two elements are necessary for the second subscripts to be placed in the order $(1,2,3,4)$.

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As I said: "each term contains one and only one element from each row and each column" means that the function that maps $j$ to $i_j$ is a permutation of $\{1,\ldots,n\}$; the exponent $k$ which is the number of transpositions that represents that permutation is equal to the sign of the permutation. So "your" definition of determinant is, word for word, the definition in the Wikipedia article. – Arturo Magidin Jun 30 '11 at 19:52
i am always thinking how the sign of the terms in the determinant are relate to the position of the determinant – Victor Jun 30 '11 at 19:59
+1 for the effort (especially compared to the previous attempt). But as I said, your definition is exactly the same as the one in Wikipedia, which explains why you have the signs you do in the cofactor expansion. Since the proof uses the exact same definition you are using, there is nothing to be done here: that is the proof that starts with "your" definition, because it's the same definition. – Arturo Magidin Jun 30 '11 at 21:07
up vote 5 down vote accepted

The proof in Wikipedia which you linked to in your previous attempt at asking this question is in fact using the very same definition of "determinant" that you are using, only with different terminology.

Wikipedia uses the definition of determinant as the sum of all terms of the form $$\mathrm{sgn}(\tau)a_{1\tau(1)}\cdots a_{n\tau(n)}$$ where $a_{ij}$ is the $(i,j)$th entry of the matrix, $\tau$ is a permutation of $\{1,2,\ldots,n\}$ (that is, a bijection $\tau\colon\{1,2,\ldots,n\}\to\{1,2,\ldots,n\}$), and $\mathrm{sgn}(\tau)$ is the sign of $\tau$.

In your description, the numbers $i_1,\ldots,i_n$ must be precisely the numbers $\{1,2,\ldots,n\}$ in some order; given a summand $$(-1)^ka_{1i_1}\cdots a_{ni_n}$$ as in your description, we define $\tau\colon\{1,2,\ldots,n\}\to\{1,2,\ldots,n\}$ by $\tau(j) = i_j$. Then $\tau$ is a permutation of $\{1,2,\ldots,n\}$, and we can express the summand as $$(-1)^ka_{1\tau(1)}\cdots a_{n\tau(n)}.$$ So the only thing left is to show that your factor $(-1)^k$ is precisely equal to $\mathrm{sgn}(\tau)$.

The sign of a permutation $\tau$ is equal to $1$ if $\tau$ can be written as a product of an even number of transpositions, and is equal to $-1$ if it can be written as a product of an odd number of transpositions. A "transposition" is precisely an "exchange of two elements".

For each "exchange" you perform to the list $i_1,i_2,\ldots,i_n$, consider the transposition that corresponds to that exchange. Performing the exchange is equivalent to composing $\tau$ with that transposition. The fact that after performing $k$ of these exchanges you get the identity tells you that $\tau$ can be written as a product of $k$ transpositions, and therefore that $\mathrm{sgn}(\tau)=(-1)^k$... exactly the same as the factor you have.

For instance, in your example $a_{13}a_{21}a_{34}a_{42}$, the permutation $\tau$ associated to this factor is, in disjoint cycle notation $\tau=(1,3,4,2)$, and in two-line notation $$\tau = \left(\begin{array}{cccc} 1&2&3&4\\ 3&1&4&2 \end{array}\right).$$ Exchanging the $1$ and $3$ is equivalent to composing $\tau$ with the transposition that exchanges $1$ and $3$ (which is $(1,3)$ in disjoint cycle notation); then exchanging $2$ and $3$; and finally exchanging $3$ and $4$, so you get $(-1)^3 = -1$. But this is the same as saying that $\tau$ is the composition of the three permutations I just mentioned: that is, that $(1,3,4,2) = (1,3)(2,3)(3,4)$, which is easy to verify. So the transposition has odd parity, hence $\mathrm{sgn}(\tau)=-1$, yielding the exact same result as yours.

Since the two definitions of determinant are in fact identical, except for the terminology used to describe the summands, the proof in Wikipedia is in fact "the proof [...] with [your] definition."

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