User:Matteo.focardi/sandbox

2020 Mathematics Subject Classification: Primary: 15Axx [MSN][ZBL]

A formula aimed at expressing the determinant of the product of two matrices $A\in\mathrm{M}_{m,n}(\mathbb{R})$ and $B\in\mathrm{M}_{n,m}(\mathbb{R})$, in terms of the sum of the products of all possible higher order minors of $A$ with corresponding minors of the same order of $B$. More precisely, if $\alpha=(1,\ldots,m)$ and $\beta$ denotes any multi-index $(\beta_1,\ldots,\beta_m)$ with $1\leq \beta_1<\ldots<\beta_m\leq n$ of length $m$, then \[ \det(AB)=\sum_\beta\det A_{\alpha\,\beta}\det B_{\beta\,\alpha}, \] where $A_{\alpha\,\beta}=(a_{\alpha_i\beta_j})$ and $B_{\beta\,\alpha}=(a_{\beta_j\alpha_i})$. In case $m>n$, no such $\beta$ exists and the right-hand side above is set to be $0$ by definition.

If $n=m$ the formula reduces to \[ \det (AB)=\det A\,\det B. \] More generally, if $A\in\mathrm{M}_{m,n}(\mathbb{R})$ and $B\in\mathrm{M}_{n,q}(\mathbb{R})$, and $N\leq\min\{m,q\}$ then any minor of order $N$ of the product matrix $AB$ can be expressed as follows by Cauchy-Binet formula \[ \det C_{\alpha,\,\gamma}=\sum_\beta\det A_{\alpha\,\beta}\det B_{\beta,\,\gamma}, \] where $\alpha=(\alpha_1\ldots,\alpha_N)$, $1\leq\alpha_1<\ldots<\alpha_N\leq m$, $\gamma=(\gamma_1,\ldots,\gamma_N)$, $1\leq\gamma_1<\ldots<\gamma_N\leq q$, and $(\beta_1,\ldots,\beta_m)$ with $1\leq \beta_1<\ldots<\beta_m\leq n$.

A number of interesting consequence of Cauchy-Binet formula is listed below. First of all, an inequality for the rank of the product matrix follows straightforwardly, i.e., \[ \mathrm{rank}(AB)\leq\min\{\mathrm{rank}A,\mathrm{rank}B\}. \] Moreover, if $m=2$, $\mathbf{a}$, $\mathbf{b}\in\mathbb{R}^n$ are two vectors, by taking $$A=\begin{pmatrix} a_{1}&\dots&a_{n}\\ b_{1}&\dots&b_{n}\\ \end{pmatrix} \quad\text{and}\quad B=\begin{pmatrix} a_{1}&b_{1}\\ \dots&\dots\\ a_{n}&b_{n}\\ \end{pmatrix} $$ by Cauchy-Binet \[ \sum_{1\leq i<j\leq n}\begin{pmatrix} a_{i}&a_{j}\\ b_{i}&b_{j}\\ \end{pmatrix}^2= \begin{pmatrix} \|\mathbf{a}\|^2&\mathbf{a}\cdot \mathbf{b}\\ \mathbf{a}\cdot \mathbf{b}&\|\mathbf{b}\|^2,\\ \end{pmatrix} \] in turn implying Cauchy-Schwartz inequality.