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Difference between revisions of "User:Matteo.focardi/sandbox"

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\det (AB)=\det A\,\det B.
 
\det (AB)=\det A\,\det B.
 
\]
 
\]
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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
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\[
 +
\det C_{\alpha,\,\gamma}=\sum_\beta\det A_{\alpha\,\beta}\det B_{\beta,\,\gamma},
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\]
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where $\alpha=(\alpha_1\ldots,\alpha_N)$, $1\leq\alpha_1<\ldots<\alpha_N\leq m$,
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$\gamma=(\gamma_1,\ldots,\gamma_N)$, $1\leq\gamma_1<\ldots<\gamma_N\leq q$, and
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$(\beta_1,\ldots,\beta_m)$ with $1\leq \beta_1<\ldots<\beta_m\leq n$.
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A number of interesting consequence of Cauchy-Binet formula is listed below.
 
A number of interesting consequence of Cauchy-Binet formula is listed below.
 
First of all, an inequality for the [[Rank|rank]] of the product matrix  
 
First of all, an inequality for the [[Rank|rank]] of the product matrix  
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\]
 
\]
 
in turn implying Cauchy-Schwartz inequality.
 
in turn implying Cauchy-Schwartz inequality.
 
If $A\in\mathrm{M}_{m,n}(\mathbb{R})$ and $B\in\mathrm{M}_{n,q}(\mathbb{R})$, in principle
 
$m\neq q$, 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
 
\[
 
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 $1\leq \beta_1<\ldots<\beta_m\leq n$.
 

Revision as of 14:46, 23 November 2012

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.

How to Cite This Entry:
Matteo.focardi/sandbox. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Matteo.focardi/sandbox&oldid=28828