Understanding Artin's derivation of the determinant formula I'm trying to understand section 1.4 of Artin. I'll mention some context so that the notation he's using is clear, even though it makes sense to me.
He defines a permutation matrix $P$ associated with permutation $p$. Then if $e_{ij}$ are "matrix units," matrices with a $ 1$ in the $ij$th place and $0$'s elsewhere, we have
$$P = e_{p(1)1} + \ldots + e_{p(n), n}.$$
So far, I understand. Then he introduces linearity of the determinant. In the $2 \times 2$ case, it gives:
\begin{align*}
\det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \det \begin{pmatrix} a & 0 \\ c & 0 \end{pmatrix} +  \det \begin{pmatrix} a & 0 \\ 0 & d \end{pmatrix} +  \det \begin{pmatrix} 0 & b \\ c & 0 \end{pmatrix} +  \det \begin{pmatrix} 0 & b \\ 0 & d \end{pmatrix}.
\end{align*}
The first and fourth matrices have a column of $0$'s, so there determinant is $0$. So far so good. He calls the remaining matrices $M$ and comments that they have exactly one entry left in each row and each column. I can see this in the $2 \times 2$ case, but don't understand why it is true, generally, though it must be obvious from linearity of the determinant.
Assuming this is true about $M$, $M$ is basically a permutation matrix, but instead of entries of $1$, they're entries $a_{ij}$ coming from some matrix $A$ whose determinant we're taking, so we have
$$M = \sum\limits_{j} a_{p(j)j} e_{p(j)j}.$$
He then states that by linearity, we have:
\begin{align*}
\det M & = (a_{p(1)1} \cdots a_{p(n)n}) \det P \\
& = (\text{sign $p$})(a_{p(1)1} \cdots a_{p(n)n}.
\end{align*}
I don't understand where either of these lines come from, and this may be my greatest source of confusion. I tried to work out a few examples by hand, but it wasn't particularly helpful.
The next observation is there is "one such term for each. permutation," so we can sum over them:
\begin{align*}
\det A & = \sum\limits_{\text{perm $p$}} (\text{sign $p$}) a_{p(1)1} \cdots a_{p(n)n}) \\
& = \sum\limits_{\text{perm $p$}} (\text{sign $p$}) a_{1p(1)} \cdots a_{np(n)})
\end{align*}
By one term for each. permutation, concretely, I assume he means that if we're looking at an $n \times n$ matrix, there are $n!$ permutations of its columns, so there are $n$ matrices $M$ (two above) and we want to sum over each of those. I believe I understand that, but I don't understand where the second, "transpose form" comes from or why it's equivalent. Is this using the fact thatthe determinant of a matrix equals the determinant of its transpose? Even if it's using that fact, I don't understand why the indices shift in this way.
I would appreciate if someone could help to clarify these.
 A: Let's observe that a determinant is a multilinear function of the rows (or columns) of a matrix.
Let's use the above fact on a 2x2 matrix $M$ given by $$M=\begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\end{bmatrix} $$ Also let's note that the unit row vectors can be written as $$e_1=(1,0),e_2=(0,1)$$ and we can write a typical row vector $r=(a, b) $ as $$r=ae_1+be_2$$ Next we can write $$\operatorname {det} (M) =\operatorname {det} \begin{bmatrix} a_{11}e_1+a_{12}e_2\\a_{21}e_1+a_{22}e_2 \end{bmatrix} $$ which further by multilinearity equals $$\operatorname {det} \begin{bmatrix} a_{11}e_1\\ a_{21}e_1 +  a_{22}e_2\end{bmatrix}+\operatorname {det} \begin{bmatrix} a_{12}e_2\\ a_{21}e_1+ a_{22}e_2\end{bmatrix} $$ This can be further split as $$\operatorname {det} \begin{bmatrix} a_{11}e_1\\ a_{21} e_1\end{bmatrix}+\operatorname {det} \begin{bmatrix} a_{11}e_1 \\ a_{22}e_2 \end{bmatrix}+\operatorname {det} \begin{bmatrix} a_{12}e_2 \\ a_{21}e_1\end{bmatrix}+\operatorname {det} \begin{bmatrix} a_{12}e_2\\ a_{22}e_2\end{bmatrix} $$ And twe can take out the matrix entries to get $$\operatorname {det} (M) =a_{11}a_{21}\operatorname {det} \begin{bmatrix} e_1\\  e_1\end{bmatrix}+a_{11}a_{22}\operatorname {det} \begin{bmatrix} e_1 \\ e_2 \end{bmatrix}+a_{12}a_{21}\operatorname {det} \begin{bmatrix} e_2 \\ e_1\end{bmatrix}+a_{12}a_{22}\operatorname {det} \begin{bmatrix} e_2\\ e_2\end{bmatrix} $$ Let's clearly notice that the determinants with at least two equal rows vanish and hence we are left with $$\operatorname {det} (M) = a_{11}a_{22}\operatorname {det} \begin{bmatrix} e_1 \\ e_2 \end{bmatrix}+a_{12}a_{21}\operatorname {det} \begin{bmatrix} e_2 \\ e_1\end{bmatrix}$$ You will thus note that we are left with only those entries which have rows with a permutation of unit row vectors (this happens in exactly the same manner with $n\times n$ matrices and we are left with exactly $n! $ determinants).
If $\sigma$ is a permutation on $n$ symbols (more formally $\sigma$ is a bijection from set $\{1,2,\dots,n\}$ to itself) then we define the sign of permutation $\sigma$ as $$\operatorname {sign} (\sigma) =\operatorname {det} \begin{bmatrix} e_{\sigma(1)}\\e_{\sigma(2)}\\ \dots\\e_{\sigma(n)} \end{bmatrix} $$ The above matrix can always be transformed into the indetity matrix $[e_1,e_2,\dots, e_n] ^{T} $ via a finite number of row exchanges and hence the sign of a permutation $\sigma$ is $(-1)^j$ where $j$ is the number of transpositions which can be applied on $\sigma$ to reduce it to identity permutation (prove that there can be multiple values of $j$ but with same parity so that $(-1)^j$ remains fixed). Using this definition one can show that $\operatorname {sign} (\sigma) =\operatorname {sign} (\sigma^{-1})$.
Coming back to our computation with the $2\times 2$ matrix $M$ we see that $$\operatorname {det} (M) =\sum_{\sigma}\operatorname {sign} (\sigma) a_{1\sigma(1)}a_{2\sigma(2)}=a_{11}a_{22}-a_{12}a_{21}$$ A similar formula holds for matrices of order $n\times n$ and if $M=[a_{ij}] _{n\times n} $ is a square matrix of order $n$ with entries $a_{ij} $ in $i$-th row and $j$-th column then we have $$\operatorname {det} (M) =\sum_{\sigma} \operatorname {sign} (\sigma) a_{1\sigma(1)}a_{2\sigma(2)}\dots a_{n\sigma(n)} $$ where $\sigma$ runs over all permutations on symbols $1,2,\dots,n$.
So you can see that we use the following properties of a determinant to prove the desired formula:

*

*it is linear function of each row (multi-linearity)

*it is $0$ if two rows are same (this is a consequence of the fact that determinant changes sign if two rows are exchanged so that it is an alternating function of row vectors).

*it takes the value $1$ on identity matrix.

The formula above in terms of permutations proves that any such function with those three properties is unique and thus determines the determinant (!!) uniquely.
Next if $\tau$ is the inverse of permutation $\sigma$ and $\sigma(i) =j$ then $\tau(j) =i$ and thus $a_{i\sigma(i)} =a_{\tau(j) j} $. Let $M=[a_{ij}]_{n\times n} $ and $M^{T} =[b_{ij}] _{n\times n} $ be the transpose of $M$ so that $b_{ij} =a_{ji} $.
Then $$\operatorname {det} (M) =\sum_{\sigma} \operatorname {sign} (\sigma) a_{1\sigma(1)}a_{2\sigma(2)}\dots a_{n\sigma(n)} $$ We can now rewrite each term in above sum with order of factors such that the second index of $a$ is in order $1,2,\dots,n$ so that $$a_{1\sigma(1)}a_{2\sigma(2)}\dots a_{n\sigma(n)}=a_{\tau(1)1}a_{\tau(2)2}\dots a_{\tau(n)n}$$ As $\sigma$ runs over all permutations on $n$ symbols so does its inverse $\tau$ and hence we get $$\operatorname {det} (M) =\sum_{\tau}\operatorname {sign} (\tau) a_{\tau(1)1}\dots a_{\tau(n) n} =\sum_{\tau} \operatorname {sign} (\tau) b_{1\tau(1)}\dots b_{n\tau(n)} =\operatorname {det} (M^T) $$

The above procedure also proves that if $f$ is any alternating and multi-linear function of the rows of $n\times n$ matrices then $$f(M) =(\operatorname {det} (M)) f(I) $$ for any square matrix $M$ of order $n$ and $I$ being the identity matrix of order $n$.
