Proof of Row Properties of the Determinant

Throughout we use the displayed form of the determinant recorded in the statement. For a real $n\times n$ matrix $B$ and $\sigma\in S_{n}$ we abbreviate the term of $B$ at $\sigma$ by $\mathrm{sgn}(\sigma)\prod_{i=1}^{n}B_{i\sigma(i)}$. Since $S_{n}$ is nonempty and finite, sums over $S_{n}$ may be evaluated through any enumeration, by Sum over a Finite Index Set.

Claim 1. The entries of $I_{n}$ are the numbers $\delta_{ij}$ of Identity Matrix. If $\sigma\in S_{n}$ and $\sigma\ne\mathrm{id}$, there is an $i\in[n]$ with $\sigma(i)\ne i$, so the factor $\delta_{i\sigma(i)}$ is $0$ and claim 4 of Properties of Finite Products gives $\prod_{i=1}^{n}\delta_{i\sigma(i)}=0$; hence the term at $\sigma$ is $0$ by Zero Products and Elementary Identities in a Field. If $\sigma=\mathrm{id}$, every factor is $\delta_{ii}=1$, so the product is $1^{n}$ by Natural Number Power of an Element of a Field, which is $1$ by claim 2 of Properties of Natural Number Powers in a Field, and the term at $\mathrm{id}$ is $\mathrm{sgn}(\mathrm{id})=1$ by claim 1 of The Sign of a Permutation is Multiplicative. Since $\{\mathrm{id}\}$ is a nonempty subset of $S_{n}$ outside which all terms vanish, claims 4 and 1 of Peeling, Splitting, and Interchange for Sums over a Finite Index Set give $\det I_{n}=1$.

Claim 2. We first record that every natural number is either $1$ or a successor: the statement $p=1$ or $p=S(p')$ for some $p'$ holds for $p=1$ and passes from $p$ to $S(p)$ trivially, so it holds for every $p$ by Principle of Induction for the Natural Numbers.

Suppose first $n=1$. By claim 2 of Basic Properties of Initial Segments of the Natural Numbers we have $[1]=\{1\}$, so every $\sigma\in S_{1}$ satisfies $\sigma(1)=1$ and $S_{1}=\{\mathrm{id}\}$. By claim 1 of Peeling, Splitting, and Interchange for Sums over a Finite Index Set, claim 1 of Properties of Finite Products and claim 1 of The Sign of a Permutation is Multiplicative, $\det B=B_{11}$ for every real $1\times1$ matrix $B$. Necessarily $i=1$, so the hypothesis reads $A_{11}=\sum_{l=1}^{m}c_{l}U_{l1}$, while $\det A^{(l)}=A^{(l)}_{11}=U_{l1}$; the assertion follows.

Now suppose $n=S(n')$ for a natural number $n'$. Fix $\sigma\in S_{n}$ and let $g_{i}:[n']\to[n]$ be the gap map of Extraction of a Term from a Finite Sum or Product in a Field at the index $i$, which by claim 1 of that lemma is a bijection onto the set of $l\in[n]$ with $l\ne i$. Put

By claim 3 of Extraction of a Term from a Finite Sum or Product in a Field, applied to the map $k\mapsto A_{k\sigma(k)}$ on $[n]$ and to the index $i$,

The matrix $A^{(l)}$ agrees with $A$ in every row other than $i$, and $g_{i}$ never takes the value $i$, so the same computation applied to $A^{(l)}$ gives

Using the hypothesis at $j=\sigma(i)$ and then claim 3 of Properties of Finite Sums twice,

Let $N$ be the number of elements of $S_{n}$ and let $\varphi:[N]\to S_{n}$ be a bijection. Evaluating the sum over $S_{n}$ through $\varphi$, interchanging the resulting double sum by Interchange of a Finite Double Sum, and using claim 3 of Properties of Finite Sums again,

and the inner sum is $\det A^{(l)}$ by Sum over a Finite Index Set.

Claim 3. By the definition of $A_{\pi}$,

By claim 4 of Permutations of an Initial Segment Form a Group under Composition the map $\rho\mapsto\rho\circ\pi$ is a bijection from $S_{n}$ onto $S_{n}$, so reindexing along it by claim 2 of Properties of a Sum over a Finite Index Set gives

Fix $\rho$ and let $a:[n]\to\mathbb{R}$ be the map $a_{k}=A_{k\rho(k)}$. Then $a_{\pi(i)}=A_{\pi(i)\rho(\pi(i))}$, so claim 2 of Invariance of Finite Sums and Products under Reindexing by a Permutation gives

By claim 2 of The Sign of a Permutation is Multiplicative, $\mathrm{sgn}(\rho\circ\pi)=\mathrm{sgn}(\rho)\mathrm{sgn}(\pi)$. Claim 4 of Properties of a Sum over a Finite Index Set now lets the constant factor $\mathrm{sgn}(\pi)$ be taken out of the sum, giving $\det A_{\pi}=\mathrm{sgn}(\pi)\det A$.

Claim 4. Let $\theta=\theta_{pq}$ be the transposition of The Sign of a Permutation is Multiplicative, an element of $S_{n}$ by claim 4 there. For every $j\in[n]$ we have $(A_{\theta})_{ij}=A_{\theta(i)j}$, which equals $A_{ij}$ when $i\notin\{p,q\}$, equals $A_{qj}=A_{pj}$ when $i=p$, and equals $A_{pj}=A_{qj}$ when $i=q$. Hence $A_{\theta}=A$. By claim 3 and by claim 4 of The Sign of a Permutation is Multiplicative,

so $(1+1)\det A=\det A+\det A=0$. By claim 8 of Elementary Order Arithmetic in an Ordered Field we have $0<1+1$, so $1+1\ne0$, and Zero Products and Elementary Identities in a Field gives $\det A=0$.

Claim 5. For every $\sigma\in S_{n}$ the factor $A_{i\sigma(i)}$ is $0$, so claim 4 of Properties of Finite Products gives $\prod_{k=1}^{n}A_{k\sigma(k)}=0$, and Zero Products and Elementary Identities in a Field makes the term at $\sigma$ equal to $0$. Thus the map summed over $S_{n}$ is the constant map $0$, which coincides with $0$ times itself, so claim 4 of Properties of a Sum over a Finite Index Set and Zero Products and Elementary Identities in a Field give $\det A=0$.

Claim 6. Fix $k\in[n]$ with $x_{k}\ne0$ and let $x_{k}^{-1}$ be its multiplicative inverse. Define an $n$-tuple $c$ of real numbers by $c_{k}=0$ and $c_{l}=-x_{k}^{-1}x_{l}$ for $l\in[n]$ with $l\ne k$.

Fix $j\in[n]$ and let $e$ and $f$ be the $n$-tuples with $e_{l}=-x_{k}^{-1}x_{l}A_{lj}$ for every $l\in[n]$, with $f_{l}=0$ for $l\ne k$ and with $f_{k}=A_{kj}$. Then $c_{l}A_{lj}=e_{l}+f_{l}$ for every $l\in[n]$: for $l\ne k$ this is the definition of $c_{l}$, and for $l=k$ the left-hand side is $c_{k}A_{kj}=0$ while the right-hand side is $e_{k}+f_{k}=-x_{k}^{-1}x_{k}A_{kj}+A_{kj}=-A_{kj}+A_{kj}=0$. By claim 2 of Properties of Finite Sums, then claim 3 of that lemma together with the hypothesis, and then claim 7 of that lemma,

So the hypothesis of claim 2 holds with $i=k$, with $m=n$ and with $U=A$. For $l\in[n]$ the matrix $A^{(l)}$ of claim 2 is then $A$ with its row $k$ replaced by the row $l$ of $A$. If $l=k$ the coefficient $c_{k}$ is $0$, so that term vanishes. If $l\ne k$, then rows $k$ and $l$ of $A^{(l)}$ both equal row $l$ of $A$, so $\det A^{(l)}=0$ by claim 4, and again the term vanishes. Every summand of $\sum_{l=1}^{n}c_{l}\det A^{(l)}$ is therefore $0$, so claim 7 of Properties of Finite Sums gives $\det A=0$.

Claim 7. By Scalar Multiple of a Real Matrix we have $(\mu A)_{i\sigma(i)}=\mu A_{i\sigma(i)}$, so claim 2 of Properties of Finite Products and Natural Number Power of an Element of a Field give

for every $\sigma\in S_{n}$. Taking the constant factor $\mu^{n}$ out of the sum over $S_{n}$ by claim 4 of Properties of a Sum over a Finite Index Set gives $\det(\mu A)=\mu^{n}\det A$.

Claim 8. By Transpose of a Real Matrix the matrix $A^{\top}$ is the real $n\times n$ matrix with $(A^{\top})_{ij}=A_{ji}$, so

By claim 4 of Permutations of an Initial Segment Form a Group under Composition the map $\rho\mapsto\rho^{-1}$ is a bijection from $S_{n}$ onto $S_{n}$, so reindexing along it by claim 2 of Properties of a Sum over a Finite Index Set gives

Fix $\rho$, put $\sigma=\rho^{-1}$ and let $a:[n]\to\mathbb{R}$ be the map $a_{k}=A_{k\rho(k)}$. Then

since $\rho\circ\rho^{-1}=\mathrm{id}$ by claim 3 of Permutations of an Initial Segment Form a Group under Composition. Claim 2 of Invariance of Finite Sums and Products under Reindexing by a Permutation therefore gives $\prod_{i=1}^{n}A_{\rho^{-1}(i)i}=\prod_{k=1}^{n}A_{k\rho(k)}$, while claim 3 of The Sign of a Permutation is Multiplicative gives $\mathrm{sgn}(\rho^{-1})=\mathrm{sgn}(\rho)$. Hence the right-hand side is $\det A$.