Proof. We first consider the case where \(\mathbf{A}\) is not invertible, then \(\det(\mathbf{A})=0\) (see the proof of Proposition 9.15). If \(\mathbf{A}\) is not invertible, then neither is \(\mathbf{A}\mathbf{B}.\) Indeed, if \(\mathbf{A}\mathbf{B}\) were invertible, then there exists a matrix \(\mathbf{C}\) such that \((\mathbf{A}\mathbf{B})\mathbf{C}=\mathbf{1}_{n}.\) But since the matrix product is associative, this also gives \(\mathbf{A}(\mathbf{B}\mathbf{C})=\mathbf{1}_{n},\) so that \(\mathbf{B}\mathbf{C}\) is a right inverse of \(\mathbf{A}.\) By Proposition 7.7, \(\mathbf{A}\) is invertible, a contradiction. Hence if \(\mathbf{A}\) is not invertible, we must also have \(\det(\mathbf{A}\mathbf{B})=0,\) which verifies that \(\det(\mathbf{A}\mathbf{B})=0=\det(\mathbf{A})\det(\mathbf{B})\) for \(\mathbf{A}\) not invertible.

If \(\mathbf{A}\) is invertible, we can write it as a product of elementary matrices and applying the second part of Lemma 9.14, we conclude that \(\det(\mathbf{A}\mathbf{B})=\det(\mathbf{A})\det(\mathbf{B}).\)

Proof. We have already seen that if \(\mathbf{A}\) is not invertible, then \(\det(\mathbf{A})=0.\) This is equivalent to saying that if \(\det(\mathbf{A})\neq 0,\) then \(\mathbf{A}\) is invertible. It thus remains to show that if \(\mathbf{A}\) is invertible, then \(\det(\mathbf{A})\neq 0.\) Suppose \(\mathbf{A}\) is invertible, then applying Proposition 10.1 gives \[\det(\mathbf{1}_{n})=\det\left(\mathbf{A}\mathbf{A}^{-1}\right)=\det(\mathbf{A})\det\left(\mathbf{A}^{-1}\right)=1\] so that \(\det(\mathbf{A})\neq 0\) and \(\det\left(\mathbf{A}^{-1}\right)=1/\det(\mathbf{A}).\)

Proof. We use induction. For \(n=1\) the condition \(A_{ij}=0\) for \(i>j\) is vacuous and (10.1) is trivially satisfied, thus the statement is anchored.

Inductive step: Assume \(n \in \mathbb{N}\) and \(n\geqslant 2.\) We want to show that if (10.1) holds for upper triangular \((n-1)\times(n-1)\)-matrices, then also for upper triangular \(n\times n\)-matrices. Let \(\mathbf{A}=(A_{ij})_{1\leqslant i,j\leqslant n} \in M_{n,n}(\mathbb{K})\) be an upper triangular matrix. Choosing \(l=1\) in the formula for \(\det(\mathbf{A}),\) we obtain \[\begin{aligned} \det(\mathbf{A})&=\sum_{k=1}^n(-1)^{k+1}A_{k1}\det\left(\mathbf{A}^{(k,1)}\right)=A_{11}\det\left(\mathbf{A}^{(1,1)}\right)+\sum_{k=2}^nA_{k1}\det\left(\mathbf{A}^{(k,1)}\right)\\ &=A_{11}\det\left(\mathbf{A}^{(1,1)}\right), \end{aligned}\] where the last equality uses that \(A_{k1}=0\) for \(k>1.\) We have \(\mathbf{A}^{(1,1)}=(A_{ij})_{2\leqslant i,j\leqslant n}\) and since \(\mathbf{A}\) is an upper triangular matrix, it follows that \(\mathbf{A}^{(1,1)}\) is an \((n-1)\times(n-1)\) upper triangular matrix as well. Hence by the induction hypothesis, we obtain \[\det(\mathbf{A}^{(1,1)})=\prod_{i=2}^nA_{ii}.\] We conclude that \(\det(\mathbf{A})=\prod_{i=1}^n A_{ii},\) as claimed.

Proof of Proposition 10.11. The matrix \(\mathbf{P}_1\) has column vectors given by \(\vec{e}_{1},\ldots,\vec{e}_n,\) hence \(\mathbf{P}_1=\mathbf{1}_{n}.\)

Using Proposition 7.2 and Theorem 7.6 it is sufficient to show that for all \(\pi,\sigma \in S_n\) we have \(f_{\mathbf{P}_{\pi\cdot \sigma}}=f_{\mathbf{P}_\pi}\circ f_{\mathbf{P}_\sigma}.\) For all \(1\leqslant i\leqslant n,\) we obtain \[f_{\mathbf{P}_\pi}\left(f_{\mathbf{P}_\sigma}(\vec{e}_i)\right)=f_{\mathbf{P}_\pi}\left(\vec{e}_{\sigma(i)}\right)=\vec{e}_{\pi(\sigma(i))}=\vec{e}_{(\pi\cdot\sigma)(i)}=f_{\mathbf{P}_{\pi\cdot \sigma}}(\vec{e}_i).\] The two maps thus agree on the ordered basis \(\mathbf{e}=(\vec{e}_1,\ldots,\vec{e}_n)\) of \(\mathbb{K}^n,\) so that the second claim follows by applying Lemma 8.6.

We have \[\mathbf{P}_{\sigma \cdot \sigma^{-1}}=\mathbf{P}_{1}=\mathbf{1}_{n}=\mathbf{P}_{\sigma}\mathbf{P}_{\sigma^{-1}}\] showing that \(\mathbf{P}_{\sigma}\) is invertible with inverse \((\mathbf{P}_{\sigma})^{-1}=\mathbf{P}_{\sigma^{-1}}.\)

Proof of Proposition 10.15. We use induction. For \(n=1\) we have \(\mathcal{X}_n=\{1\}\) and the only permutation is the identity permutation \(1,\) so the statement is trivially true and hence anchored.

Inductive step: Assume \(n \in \mathbb{N}\) and \(n\geqslant 2.\) We want to show that if the claim holds for \(S_{n-1},\) then also for \(S_n.\) Let \(\sigma \in S_n\) and define \(k=\sigma(n).\) Then the permutation \(\sigma_1=\tau_{n,k}\sigma\) satisfies \(\sigma_1(n)=\tau_{n,k}\sigma(n)=\tau_{n,k}(k)=n\) and hence does not permute \(n.\) Restricting \(\sigma_1\) to the first \(n-1\) elements, we obtain a permutation of \(\{1,\ldots,n-1\}.\) By the induction hypothesis, we thus have \({\tilde{m}} \in \mathbb{N}\) and \(\tau_{k_1,l_1},\ldots \tau_{k_{{\tilde{m}}}},\tau_{l_{{\tilde{m}}}} \in S_n\) such that \[\sigma_1=\tau_{k_{{\tilde{m}}},l_{{\tilde{m}}}}\cdots \tau_{k_1,l_1}=\tau_{n,k}\sigma.\] Since \(\tau_{n,k}^2=1,\) multiplying from the left with \(\tau_{n,k}\) gives \(\sigma=\tau_{n,k}\tau_{k_{{\tilde{m}}},l_{{\tilde{m}}}}\cdots \tau_{k_1,l_1},\) the claim follows with \(m={\tilde{m}}+1.\)

Sketch of a proof. We can define a function \(f : \mathbb{K}^n \to \mathbb{K},\) \(\vec{x} \mapsto \det(\mathbf{V}_{\vec{x}}).\) By the Leibniz formula, the function \(f\) is a polynomial in the variables \(x_i\) with integer coefficients. If we freeze all variables of \(f\) except the \(\ell\)-th variable, then we obtain a function \(g_{\ell} : \mathbb{K}\to \mathbb{K}\) of one variable \(x_{\ell}.\) For \(1\leqslant i\leqslant n\) with \(i\neq \ell\) we have \(g_{\ell}(x_i)=0,\) since we compute the determinant of a matrix with two identical rows, the \(\ell\)-th row and the \(i\)-th row. Factoring the zeros, we can thus write \(g_{\ell}(x_{\ell})=q_{\ell}(x_{\ell})\prod_{1\leqslant i\leqslant n, i\neq \ell}(x_{\ell}-x_i)\) for some polynomial \(q_{\ell}.\) We can repeat this argument for all \(\ell\) and hence can write \(\det(\mathbf{V}_{\vec{x}})=q(\vec{x})\prod_{1\leqslant i<j\leqslant n}(x_j-x_i)\) for some polynomial \(q(\vec{x}).\)

On the other hand, if we multiply all the \(x_i\) by a constant \(s \in \mathbb{K},\) the determinant multiplies by \(s^{1 + 2 + \dots + (n-1)} = s^{n(n-1)/2}.\) The product \(\prod_{1\leqslant i<j\leqslant n}(x_j-x_i)\) has the same scaling behaviour (since it has \(n(n-1)/2\) linear factors); so \(q\) must be invariant under scaling the variables. This implies that \(q\) has to be constant.

Using the Leibniz formula, we see that the summand of \(\det(\mathbf{V}_{\vec{x}})\) corresponding to the identity permutation is the product of the diagonal entries of \(\mathbf{V}_{\vec{x}},\) that is, \(x_2(x_3)^2\cdots(x_n)^{n-1}\) (and no other term in the sum has this combination of exponents). Taking the first term in all factors of \(\prod_{1\leqslant i<j\leqslant n}(x_j-x_i),\) we also obtain \(x_2(x_3)^2\cdots(x_n)^{n-1}\) (and no other term in the product has these exponents). Hence the constant \(q\) must be 1, and so \(\det(\mathbf{V}_{\vec{x}})=\prod_{1\leqslant i<j\leqslant n}(x_j-x_i),\) as claimed.

Proof. Let \(\mathbf{A}=(A_{ij})_{1\leqslant i,j\leqslant n}.\) For \(1\leqslant i\leqslant n\) we obtain for the \(i\)-th diagonal entry \[[\operatorname{Adj}(\mathbf{A})\mathbf{A}]_{ii}=\sum_{k=1}^n(-1)^{i+k}\det\left(\mathbf{A}^{(k,i)}\right)A_{ki}=\det(\mathbf{A}),\] where we use the Laplace expansion (9.5) of the determinant. The diagonal entries of \(\operatorname{Adj}(\mathbf{A})\mathbf{A}\) are thus all equal to \(\det \mathbf{A}.\) For \(1\leqslant i,j\leqslant n\) with \(i\neq j\) we have \[[\operatorname{Adj}(\mathbf{A})\mathbf{A}]_{ij}=\sum_{k=1}^n(-1)^{i+k}\left(\det \mathbf{A}^{(k,i)}\right)A_{kj}.\] We would like to interpret this last expression as a Laplace expansion. We consider a new matrix \(\hat{\mathbf{A}}=(\hat{A}_{ij})_{1\leqslant i,j\leqslant n}\) which is identical to \(\mathbf{A},\) except that the \(i\)-th column of \(\mathbf{A}\) is replaced with the \(j\)-th column of \(\mathbf{A},\) that is, for \(1\leqslant k\leqslant n,\) we have \[\tag{10.4} \hat{A}_{kl}=\left\{\begin{array}{cc} A_{kj}, & l=i,\\ A_{kl}, & l\neq i.\end{array}\right.\] Then, for all \(1\leqslant k \leqslant n\) we have \(\hat{\mathbf{A}}^{(k,i)}=\mathbf{A}^{(k,i)},\) since the only column in which \(\mathbf{A}\) and \(\hat{\mathbf{A}}\) are different is removed in \(\mathbf{A}^{(k,i)}.\) Using (10.4), the Laplace expansion of \(\hat{\mathbf{A}}\) with respect to the \(i\)-th column gives \[\begin{aligned} \det\hat{\mathbf{A}}&=\sum_{k=1}^n(-1)^{(i+k)}\hat{A}_{ki}\det\left(\hat{\mathbf{A}}^{(k,i)}\right)=\sum_{k=1}^n(-1)^{i+k}\left(\det \mathbf{A}^{(k,i)}\right)A_{kj}\\ &=[\operatorname{Adj}(\mathbf{A})\mathbf{A}]_{ij} \end{aligned}\] The matrix \(\hat{\mathbf{A}}\) has a double occurrence of the \(i\)-th column, hence its column vectors are linearly dependent. Therefore \(\hat{\mathbf{A}}\) is not invertible by Proposition 3.18 and so \(\det\hat{\mathbf{A}}=[\operatorname{Adj}(\mathbf{A})\mathbf{A}]_{ij}=0\) by Corollary 10.2. The off-diagonal entries of \(\operatorname{Adj}(\mathbf{A})\mathbf{A}\) are thus all zero and we conclude \(\operatorname{Adj}(\mathbf{A})\mathbf{A}=\det(\mathbf{A})\mathbf{1}_{n}.\) Using the row version of the Laplace expansion we can conclude analogously that \(\mathbf{A}\operatorname{Adj}(\mathbf{A})=\det(\mathbf{A})\mathbf{1}_{n}.\)

Finally, if \(\mathbf{A}\) is invertible, then \(\det \mathbf{A}\neq 0\) by Corollary 10.2, so that \(\mathbf{A}^{-1}=\operatorname{Adj}(\mathbf{A})/\det(\mathbf{A}),\) as claimed.