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If $(j_1,j_2,\dots, j_n)$ is an ordered $n$-tuple of integers, define $$s(j_1, j_2, \dots, j_n)=\prod \limits_{p<q}\text{sgn}(j_q-j_p).$$ Let $[A]$ be the matrix of a linear operator on $\mathbb{R}^n$, relative to the standard basis $\{e_1, e_2, \dots, e_n\}$, with entries $a(i,j)$ in the $i$th row and $j$th column. The determinant of $[A]$ is defined to be the number $$\det[A]=\sum s(j_1, j_2, \dots, j_n)a(1,j_1)a(2,j_2)\dots a(n,j_n)$$ where the sum extends over all permutations $(j_1,j_2,\dots, j_n)$ of $(1,2,\dots,n)$. Note that this sum contains $n!$ terms.

Theorem: If $[A]_1$ is obtained from $[A]$ by interchanging two columns, then $\det [A]_1=-\det [A].$

Proof: It is an immediate consequence of the fact that $s(j_1,j_2,\dots,j_n)$ changes sign if any two of the $j$'s are interchanged.

This is definition of determinant from Rudin's book. I understood the first part of text but I can't understand proof of theorem about 3 days :(

For example, suppose we have $3\times 3$ matrix $[A]$ with entries $a(i,j)$ and interchanging the first two columns we get a new matrix: $$[A]_1=\begin{bmatrix} a_{12} & a_{11} & a_{13} \\ a_{22} & a_{21} & a_{23} \\ a_{32} & a_{31} & a_{33} \end{bmatrix}$$

Then by definition of determinant we have: $$\det[A]_1=\sum \limits_{(k_1,k_2,k_3)}s(k_1,k_2,k_3)a_{1k_1}a_{2k_2}a_{3k_3}$$ and it's clear that $\det[A]=\det[A]_1$ and it doesn't change sign.

Sorry if this answer is repeated but I want to prove it only using above definitons. I would be very thankful for any answer!

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2 Answers 2

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The confusion probably stems from your retaining the original labelling of the entries of the matrix. For instance, in the formula for $\det [A]_1$, $a_{11}$ would refer to the first entry of $[A]_1$ which is actually called $a_{12}$. You probably conflated the two when you claimed the determinant did not change. Let $b_{ij}$ denote the entries of $[A]_1$ and note that for example $b_{x1}=a_{x2}$ and and so on.

To show the determinant changes sign, it suffices to show that each addend in the formula of $\det [A]$ is the negative of a corresponding addend of $\det [A]_1$. (For instance, $s(3,2,1) a_{13} a_{22} a_{31}$ in the definition of $\det [A]$ corresponds with $s(3,1,2) b_{13} b_{21}b_{32}=-s(3,2,1) a_{13} a_{22}a_{31}$.)

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  • $\begingroup$ Mmm. Right! But how to generalize it to $n\times n$ matrices? I guees that it would be really difficult. $\endgroup$
    – RFZ
    Feb 2, 2016 at 7:27
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Your

Then by definition of determinant we have: $$\det[A]_1=\sum \limits_{(k_1,k_2,k_3)}s(k_1,k_2,k_3)a_{1k_1}a_{2k_2}a_{3k_3}$$

is wrong. It should be

$$\det[A]_1=\sum \limits_{(k_1,k_2,k_3)}s(k_1,k_2,k_3)a^1_{1k_1}a^1_{2k_2}a^1_{3k_3}$$

where $a^1_{ij}$ are the entries of the permuted matrix. These satisfy $a^1_{i2}=a_{i1}$ and $a^1_{i1}=a_{i2}$. Plugging this in gives does not give you what you claimed. Instead you need to play aorund a bit and see that the switch of columns essentially switches one of the terms in each $s$, resulting in a sign switch of the determinant.

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  • $\begingroup$ I understood your remark! I checked it in paper and I understand my mistake. But I guess that for $n\times n$ matrices it would be hard. $\endgroup$
    – RFZ
    Feb 2, 2016 at 7:34
  • $\begingroup$ It is actually not if you have the right (abstract enough) definitions. Check other books to get succinct proofs (and more importantly ways to introduce the determinant) that don't use indices and writing down matrices with their elements $\endgroup$
    – Bananach
    Feb 2, 2016 at 7:48
  • $\begingroup$ I looked other books but their proof liitle bit hard to me because they need knowledge about permutations, transpositions and their signes. Can you help me to prove that theorem only using Rudin's notations? $\endgroup$
    – RFZ
    Feb 2, 2016 at 7:51

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