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I have a question about the second paragraph of the following excerpt, which is taken from p.115 of Introductory Mathematics: Algebra and Analysis by Geoff Smith:

In the light of these properties of determinant, we calculate $|X|$ again where $X=\left(x_{i j}\right)$ is a 3 by 3 matrix. Instead of using Definition $4.7$, let us see how far we get in an attempt to evaluate $|X|$ using Remark 4.3.

Think of $X$ as a column vector, its entries being row vectors. Let the $i$-th row be $\mathbf{x}_{i}$. Now let $\mathbf{e}_{i}$ be the $i$-th row of $I_{n}$, so that $\mathbf{x}_{i}=\sum_{j} x_{i j} \mathbf{e}_{j}$. We now use property (c) in Remark $4.3$ (determinant being a multilinear function) to find that $|X|$ is $$ \begin{aligned} x_{11} x_{22} x_{33}\left|\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right|+x_{11} x_{23} x_{32}\left|\begin{array}{lll} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right|\\ +x_{12} x_{23} x_{31}\left|\begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}\right|+x_{12} x_{21} x_{33}\left|\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right|\\ +x_{13} x_{21} x_{32}\left|\begin{array}{lll} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right|+x_{13} x_{22} x_{31}\left|\begin{array}{lll} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}\right| \end{aligned} $$

Specifically, my question is what is the connection between the second paragraph and the formula below it. How do the authors get from

Now let $\mathbf{e}_{i}$ be the $i$-th row of $I_{n}$, so that $\mathbf{x}_{i}=\sum_{j} x_{i j} \mathbf{e}_{j}$.

to the expression for the determinant of $X$?

Extra (the whole book: https://docdro.id/gnu93vl):

Definition 4.7 Suppose that $A=\left(a_{i j}\right)$ is an $n$ by $n$ matrix. We define the determinant $|A|$ inductively. Thus we assume that we know how to calculate a determinant $|X|$ where $X$ is an $n-1$ by $n-1$ matrix, and start the induction off by defining the determinant of a 1 by 1 matrix $(a)$ to be $a$.

Pick any row of $A$, say the $i$-th row. We will work out $|A|$ by using the entries of the $i$-th row of $A$ and the determinants of some $n-1$ by $n-1$ matrices. For each entry $a_{i j}$ in the $i$-th row, let $A_{i j}$ be the $n-1$ by $n-1$ matrix obtained by striking out the $i$-th row and $j$-th column of $A$. Let $c_{i j}=(-1)^{i+j}\left|A_{i j}\right|$. Now let $|A|=\sum_{j=1}^{n} a_{i j} c_{i j}$.

Remark 4.3 (c) The map is multilinear in each of the $n$ row vector variables. In other words, if you choose $i$ in the range $1 \leq i \leq n$ and keep all arguments fixed except the variable in position $i$, you obtain a linear map from $\mathbb{R}^{n}$ to $\mathbb{R}$. To be explicit, if we fix all rows except one, determinant defines a function $\alpha$ from $\mathbb{R}^{n}$ to $\mathbb{R}$ by varying the distinguished row. To say that $\alpha$ is linear is to assert that $$ \alpha(\lambda \mathbf{x}+\mu \mathbf{y})=\lambda \alpha(\mathbf{x})+\mu \alpha(\mathbf{y}) \forall \lambda, \mu \in \mathbb{R}, \forall \mathbf{x}, \mathbf{y} \in \mathbb{R}^{n} $$

user1551
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Treex
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    Remark 4.3, definition 4.7,... in which book? – FShrike Jun 27 '22 at 16:13
  • see the Wikipedia article on the Leibniz determinant formula, anyway. It's terse but it should cover everything – FShrike Jun 27 '22 at 16:14
  • @FShrike Introductory Mathematics: Algebra and Analysis, page 115 – Treex Jun 28 '22 at 14:02
  • Tell the post that, rather than me specifically. My point was, it’s hard to answer your question unless we have a copy of this book to hand. Try to add everything relevant, including their proof up to where you got stuck – FShrike Jun 28 '22 at 14:04
  • @FShrike I agree, it could indeed be harder to answer. I tried to keep the post as short as possible and hence omitted some parts, but nevertheless I now added the missing text as well as shared the link to the book. – Treex Jun 28 '22 at 14:27
  • The matrix addition you see is the result of applying the multi-linearity of the determinant. These properties are independent of how the determinant is defined. Do you understand the difference between linearity and multi-linearity? – John Douma Jun 28 '22 at 22:24
  • @JohnDouma I suppose I do, here is a quite nice explanation: https://math.stackexchange.com/questions/1403735/why-is-determinant-a-multilinear-function but I still don't understand how the authors derived the above. Perhaps you have an idea? – Treex Jun 29 '22 at 06:12
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    @Treex see my answer. Hope you will understand. – Sourav Ghosh Jul 11 '22 at 17:03
  • @LostinSpace Perfect, thank you! That's what I was looking for. – Treex Jul 12 '22 at 09:15

1 Answers1

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Let us illustrate by examples.

${{\color{red}{2×2 \space Matrix }}}$

Let $A=\begin{pmatrix} x_{11}&x_{12}\\x_{21}&x_{22}\end{pmatrix}$

$\begin{align}\color{blue}{\text{Row} \space 1} : x_1&=\begin{bmatrix}x_{11} &x_{12}\end{bmatrix}\\&=x_{11}\begin{bmatrix} 1 & 0 \end{bmatrix} + x_{12}\begin{bmatrix} 0 & 1 \end{bmatrix} \\&=x_{11}e_1+x_{12}e_{2}\end{align}$

$\begin{align}\color{green}{\text{Row} \space 2} : x_2&=\begin{bmatrix}x_{21} &x_{22}\end{bmatrix}\\&=x_{21}\begin{bmatrix} 1 & 0 \end{bmatrix} + x_{22}\begin{bmatrix} 0 & 1 \end{bmatrix} \\&=x_{21}e_1+x_{22}e_{2}\end{align}$

$\begin{align}\begin{vmatrix}\color{blue}{\text{Row}1}\\\color{green}{\text{Row }2}\end{vmatrix}&= \begin{vmatrix} x_{11}e_1+x_{12}e_{2} \\\color{green}{\text{Row }2}\end{vmatrix}\\& {\overset{\color{red}1}{=}}\begin{vmatrix} x_{11}e_1 \\\color{green}{\text{Row }2}\end{vmatrix} +\begin{vmatrix} x_{12}e_{2} \\\color{green}{\text{Row }2}\end{vmatrix}\\&{\overset{\color{red}2}{=}}x_{11}\begin{vmatrix} e_1 \\x_{21}e_1+x_{22}e_2\end{vmatrix} +x_{12}\begin{vmatrix} e_{2} \\x_{21}e_1+x_{22}e_2\end{vmatrix} \\&{\overset{\color{red}3}{=}}x_{11}x_{21} \begin{vmatrix} e_1\\e_1\end{vmatrix}+x_{11}x_{22} \begin{vmatrix} e_1\\e_2\end{vmatrix}+ x_{11}x_{21} \begin{vmatrix} e_2\\e_1\end{vmatrix}+x_{12}x_{22} \begin{vmatrix} e_2\\e_2\end{vmatrix}\\& {\overset{\color{red}4}{=}} x_{11}x_{21} \cdot 0 +x_{11}x_{22} \cdot 1+ x_{12}x_{21} \cdot (-1)+x_{12}x_{22} \cdot 0 \\&= x_{11}x_{22}-x_{12}x_{21} \end{align}$

$\begin{align}\begin{vmatrix}x_{11}&x_{12}\\x_{21}&x_{22}\end{vmatrix}&=\begin{vmatrix}x_{11}&0\\x_{21}&x_{22}\end{vmatrix}+ \begin{vmatrix}0&x_{12}\\x_{21}&x_{22}\end{vmatrix}\\&=\begin{vmatrix}x_{11}&0\\x_{21}&0\end{vmatrix}+ \begin{vmatrix}x_{11}&0\\0&x_{22}\end{vmatrix}+\begin{vmatrix}0&x_{12}\\x_{21}&0\end{vmatrix}+\begin{vmatrix}0&x_{12}\\0&x_{22}\end{vmatrix}\\&=0+x_{11}x_{22} \begin{vmatrix}1&0\\0&1\end{vmatrix}+x_{12}x_{21} \begin{vmatrix}0&1\\1&0\end{vmatrix}+0\\&=x_{11}x_{22}-x_{12}x_{21} \end{align}$

${\overset{\color{red}1}{=}}:\text{Additivity in Row }1$

${\overset{\color{red}2}{=}}:\text{Homogeneity in Row }1$

${\overset{\color{red}3}{=}}:\text{Linearity in Row }2$

${\overset{\color{red}4}{=}}:\begin{cases}\text{Two equal Row }:\det(A) =0 \\\text{Two Row exchange }:\det(A') =-\det(A)\end{cases}$

Extension for $3×3$ matrix is similar.

Let $A=\begin{pmatrix} x_{11}&x_{12}&x_{13}\\x_{21}&x_{22}&x_{23}\\x_{31}&x_{32}&x_{33}\end{pmatrix}$

$\begin{align}\color{blue}{\text{Row} \space 1} : x_1&=\begin{bmatrix}x_{11} &x_{12}&x_{13}\end{bmatrix}\\&=x_{11}\begin{bmatrix} 1 & 0 &0\end{bmatrix} + x_{12}\begin{bmatrix} 0 & 1 &0\end{bmatrix}+x_{13}\begin{bmatrix}0&0&1 \end{bmatrix} \\&=x_{11}e_1+x_{12}e_{2}+x_{13}e_3\end{align}$

$\begin{align}\color{green}{\text{Row} \space 2} : x_2&=\begin{bmatrix}x_{21} &x_{22}&x_{23}\end{bmatrix}\\&=x_{21}\begin{bmatrix} 1 & 0 &0\end{bmatrix} + x_{22}\begin{bmatrix} 0 & 1 &0\end{bmatrix}+x_{23}\begin{bmatrix}0&0&1 \end{bmatrix} \\&=x_{21}e_1+x_{22}e_{2}+x_{23}e_3\end{align}$

$\begin{align}\color{brown}{\text{Row} \space 3} : x_3&=\begin{bmatrix}x_{31} &x_{32}&x_{33}\end{bmatrix}\\&=x_{31}\begin{bmatrix} 1 & 0 &0\end{bmatrix} + x_{32}\begin{bmatrix} 0 & 1 &0\end{bmatrix}+x_{33}\begin{bmatrix}0&0&1 \end{bmatrix} \\&=x_{31}e_1+x_{32}e_{2}+x_{33}e_3\end{align}$

Now expand linearly and compute $3! $ determinants .

Sourav Ghosh
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