Brief moment in theorem about determinant from baby Rudin

Question

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!

Answer 1

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.

Answer 2

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 $s(3,1,2) b_{13} b_{21}b_{32}=-s(3,2,1) a_{13} a_{22}a_{31}$.)