# Proof of the Laplace formula using alternating tensors.

I am reading "Differential Forms with applications to the physical sciences" of Harley Flanders. I came across the proof of the Laplace formula, there were some details left to the reader, I would like to know if my proof is correct. Let $$\{a_{i,\,j}\}_{i=1,\dots,n,\,j=1\dots n}$$ be an $$n\times n$$ matrix. For an order $$p-$$tuple $$H=\{h_{1},\dots,\,h_{p}\}$$ , set $$b_{H}=\begin{vmatrix} a_{1,\,h_{1}}&& \dots && a_{1,\,h_{p}}\\ \vdots && && \vdots \\ a_{p,h_{1}} && \dots && a_{p,h_{p}} \end{vmatrix}.$$ Set $$p+q=n.$$ For $$K=\{k_{1},\,\dots\, k_{q}\},$$ set $$c_{K}=\begin{vmatrix} a_{p+1,\,k_{1}}&& \dots && a_{1,\,k_{q}}\\ \vdots && && \vdots \\ a_{n,k_{1}} && \dots && a_{n,k_{q}} \end{vmatrix}.$$ I will indicate with $$H^{c}$$ the ordered $$q-$$tuple complementar of $$H$$. Set $$\alpha_{i}=\sum_{j=1}^{n}a_{i,\,j}\,\sigma^{j},$$ where $$(\sigma^{j})$$ is a base of $$\mathbb{R}^{n}.$$ First we prove by induction that $$\alpha_{1}\wedge\,\dots\,\wedge \alpha_{p}=\sum_{H} b_{H}\,\sigma^{H}$$ and $$\alpha_{p+1}\wedge\,\dots\, \wedge \alpha_{n}=\sum_{K}c_{K}\,\sigma^{K}.$$ The case $$p=1$$ is trivial. If $$p>1$$ we have (I am not really sure about the third equality) \begin{align*} \alpha_{1}\wedge\,\dots\,\wedge \alpha_{p} =\left(\sum_{j=1}^{n} a_{j,1}\, \sigma^{j}\right)\wedge \, \left( \alpha_{2}\,\wedge \dots \, \wedge \alpha_{p} \right)\\ =\sum_{j=1}^{n}\,a_{j,1}\,\sigma^{j} \wedge \sum_{\bar{H}}b_{\bar{H}}\,\sigma^{\bar{H}}=\sum_{i=1}^{n}a_{h_{i},1}\,\sigma^{h_{i}}\wedge b_{\hat{H}_{i}}\, \sigma^{\hat{H}_{i}}\\ =\sum_{i=1}^{n} a_{1,h_{i}}\,b_{\hat{H}_{i}}\,(-1)^{h_{i}-1}\,\sigma^{H}=\sum_{H}\, b_{H}\,\sigma^{H}. \end{align*} Where the last equality is by the definition of the determinant. Remark: the notation $$\hat{H}_{i}$$ means the ordered $$p-1-$$tuple obtained by $$H$$ subtracting $$h_{i}.$$

The other formula $$\alpha_{p+1}\wedge\,\dots\, \wedge \alpha_{n}=\sum_{K}c_{K}\,\sigma^{K}$$ is proven similarly. So the Laplace formula is readily proven : \begin{align*} \det(A)\,\sigma^{1}\wedge \dots \wedge \sigma^{n}=\left(\alpha_{1}\wedge \dots \wedge \alpha_{p}\right)\wedge \left( \alpha_{p+1}\wedge \dots \alpha_{n} \right)=\sum_{H,K} b_{H}\,c_{K}\,\sigma^{H}\wedge \sigma^{K} \end{align*} But $$\sigma^{H}\wedge \sigma^{K}=\begin{cases} 0 \quad K\neq H^{c}\\ sgn(H,H^{c}) \quad K=H^{c} \end{cases}$$ In conclusion \begin{align*} \det(A)\,\sigma^{1}\wedge \dots \wedge \sigma^{n}=\sum_{H,K} b_{H}\,c_{K}\,\sigma^{H}\wedge \sigma^{K}=\left( \sum_{H} sgn(H,H') a_{H}\,b_{H'} \right) \cdot \sigma_{1}\wedge \dots \sigma_{n} \end{align*} Remark: obviously the notation $$sgn(H,H')$$ stands for the sign of the permutation $$\begin{pmatrix} 1 &2 & \cdots &p& p+1 & \dots & n\\ h_{1} & h_{2} & \cdots & h_{p}& k_{1} & \cdots &k_{q}. \end{pmatrix}$$

• I tried this book and felt it sucked there are far superior books on this topicc Commented Aug 29, 2022 at 7:46
• @TrystwithFreedom so far I have liked it, but I am open to suggestions. Commented Aug 29, 2022 at 7:52
• At quick glance it looks like you have typos. In the definition of $c_K$, I think you want $k$'s not $h$'s. In the definition of $\alpha_i$, I think you want $a_{i,j}$ not $\alpha_{i,j}$. I didn't read any further. I'd recommend proofreading if you want someone to review your work carefully. Commented Sep 2, 2022 at 14:41

First, it appears you are trying to prove the classical Laplace expansion formula for an $$n\times n$$ real matrix $$A=(a_{ij})$$ along the first $$p$$ rows, which can be written more explicitly as $$\det A=\sum_{h_1<\cdots where $$h_{p+1}<\cdots is the complementary ordered $$(n-p)$$-tuple of $$h_1<\cdots, $$\mathrm{sgn}(h_1,\ldots,h_p)=(-1)^{\sum_{i=1}^p(h_i-i)}\tag{2}$$ and $$A_{i_1\cdots i_k}^{j_1\cdots j_k}$$ denotes the $$k\times k$$ minor determinant of $$A$$ from rows $$i_1<\cdots and columns $$j_1<\cdots. (Note (1) is equivalent to your final outer equation, just using slightly more explicit notation; in particular, (2) is the sign of the permutation you describe.)
As in your attempted proof we can let $$\sigma^1,\ldots,\sigma^n$$ be a basis of $$\mathbb{R}^n$$ and $$\alpha_i=\sum_{j=1}^n a_{ij}\sigma^j$$.
Your attempted inductive proof of the expansion for $$\alpha_1\wedge\cdots\wedge\alpha_p$$ is problematic, even ignoring the typos. The right-hand side of your third equality doesn't make sense since $$H$$ (and hence $$h_i$$ and $$\hat{H}_i$$) is not defined there. However, it is true that $$\alpha_1\wedge\cdots\wedge\alpha_p=\sum_{h_1<\cdots where $$\sigma^{h_1\cdots h_p}=\sigma^{h_1}\wedge\cdots\wedge\sigma^{h_p}$$, and this can be proved inductively using the Laplace cofactor expansion along a single row of a matrix (a simpler result which can be proved independently and which I assume you're familiar with).
As an illustrative example if $$n=p=3$$, then $$\alpha_2\wedge\alpha_3=\begin{vmatrix}a_{21}&a_{22}\\a_{31}&a_{32}\end{vmatrix}\sigma^{12}+\begin{vmatrix}a_{21}&a_{23}\\a_{31}&a_{33}\end{vmatrix}\sigma^{13}+\begin{vmatrix}a_{22}&a_{23}\\a_{32}&a_{33}\end{vmatrix}\sigma^{23}$$ so it follows that \begin{align*} \alpha_1\wedge\alpha_2\wedge\alpha_3&=\Biggl(a_{11}\begin{vmatrix}a_{22}&a_{23}\\a_{32}&a_{33}\end{vmatrix}-a_{12}\begin{vmatrix}a_{21}&a_{23}\\a_{31}&a_{33}\end{vmatrix}+a_{13}\begin{vmatrix}a_{21}&a_{22}\\a_{31}&a_{32}\end{vmatrix}\Biggr)\sigma^{123}\\ &=\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix}\sigma^{123} \end{align*} I trust you can generalize the reasoning here.