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

#### Peter

##### Well-known member
MHB Site Helper
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: 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
$$\displaystyle = \sum_{ \sigma } a_{ \sigma (1) 1 } a_{ \sigma (2) 2 } \ \text{det} ( e_{ \sigma (1) }, e_{ \sigma (2) } )$$

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
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