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Permutations and Determinants ... Walschap, Theorem 1.3.1 ... ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,881
Hobart, Tasmania
I am reading Gerard Walschap's book: "Multivariable Calculus and Differential Geometry" and am focused on Chapter 1: Euclidean Space ... ...

I need help with an aspect of the proof of Theorem 1.3.1 ...

The start of Theorem 1.3.1 and its proof read as follows:



Walschap - Theorem 1.3.1 ... .png



I tried to understand how/why

\(\displaystyle \text{det} ( v_1, \cdot \cdot \cdot , v_n ) = \sum_{ \sigma } a_{ \sigma (1) 1 } , \cdot \cdot \cdot , a_{ \sigma (n) n } \ \text{det} ( e_{ \sigma (1) } , \cdot \cdot \cdot , e_{ \sigma (n) } \)


where the sum runs over all maps \(\displaystyle \sigma \ : \ J_n = \{ 1 , \cdot \cdot \cdot , n \} \to J_n = \{ 1 , \cdot \cdot \cdot , n \}\)

... so ...

... I tried an example with \(\displaystyle \text{det} \ : \ (\mathbb{R}^n)^n \to \mathbb{R}\) ... ...


so we have \(\displaystyle v_1 = \sum_k a_{ k1 } e_k = a_{11} e_1 + a_{21} e_2\)


and


\(\displaystyle v_2 = \sum_k a_{ k2 } e_k = a_{12} e_1 + a_{22} e_2 \)


and then we have


\(\displaystyle \text{det} ( v_1, v_2 )\)


\(\displaystyle = \text{det} ( a_{11} e_1 + a_{21} e_2 , a_{12} e_1 + a_{22} e_2 )\)


\(\displaystyle = a_{11} a_{12} \ \text{det} ( e_1, e_1 ) + a_{11} a_{22} \ \text{det} ( e_1, e_2 ) + a_{21} a_{12} \ \text{det} ( e_2, e_1 ) + a_{21} a_{22} \ \text{det} ( e_2, e_2 )\)


\(\displaystyle = \sum_{ \sigma } a_{ \sigma (1) 1 } a_{ \sigma (2) 2 } \ \text{det} ( e_{ \sigma (1) }, e_{ \sigma (2) } ) \)


where the sum runs over all maps


\(\displaystyle \sigma \ : \ J_n = \{ 1, 2 \} \to J_n = \{ 1 , 2 \}\) ...


... ... that is the sum runs over the two permutations


\(\displaystyle \sigma_1 = \begin{bmatrix} 1 & 2 \\ 1 & 2 \end{bmatrix}\) and \(\displaystyle \sigma_2 = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\)


BUT how does the formula \(\displaystyle \sum_{ \sigma } a_{ \sigma (1) 1 } a_{ \sigma (2) 2 } \ \text{det} ( e_{ \sigma (1) }, e_{ \sigma (2) } \) ...

... incorporate or deal with the terms involving \(\displaystyle \text{det} ( e_1, e_1 )\) and \(\displaystyle \text{det} ( e_2, e_2 )\) ... ... ?


Hope someone can help ...?

Peter
 
Last edited:

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,483
\(\displaystyle = \sum_{ \sigma } a_{ \sigma (1) 1 } a_{ \sigma (2) 2 } \ \text{det} ( e_{ \sigma (1) }, e_{ \sigma (2) } ) \)


where the sum runs over all maps


\(\displaystyle \sigma \ : \ J_n = \{ 1, 2 \} \to J_n = \{ 1 , 2 \}\) ...


... ... that is the sum runs over the two permutations


\(\displaystyle \sigma_1 = \begin{bmatrix} 1 & 2 \\ 1 & 2 \end{bmatrix}\) and \(\displaystyle \sigma_2 = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\)
Permutations are not all maps from {1, 2} to {1, 2}. When we are expanding the determinant in the first display equation in the picture, we take one term $a_{jk}e_j$ from each argument, i.e., from each sum $a_{ik}e_1+\dots+a_{nk}e_n$. The function $\sigma$ maps the number of the argument $k$ to the number of the term $j$ we select from that argument. The expansion includes all possible variants of choosing terms from arguments of det, so the sum ranges over all functions, not just bijections. But $\sigma$ that is not a bijection leads to a zero determinant.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,881
Hobart, Tasmania
Permutations are not all maps from {1, 2} to {1, 2}. When we are expanding the determinant in the first display equation in the picture, we take one term $a_{jk}e_j$ from each argument, i.e., from each sum $a_{ik}e_1+\dots+a_{nk}e_n$. The function $\sigma$ maps the number of the argument $k$ to the number of the term $j$ we select from that argument. The expansion includes all possible variants of choosing terms from arguments of det, so the sum ranges over all functions, not just bijections. But $\sigma$ that is not a bijection leads to a zero determinant.


Thanks Evgeny ...

Still reflecting/puzzling over how exactly \(\displaystyle \sigma\) is defined as a function ...

Peter