Difference between revisions of "User:Matteo.focardi/sandbox"
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[[Multiindex|multi-index]] $(k_1,\ldots,k_m)$ with $1\leq k_1<\ldots<k_m\leq n$ of length $m$, then | [[Multiindex|multi-index]] $(k_1,\ldots,k_m)$ with $1\leq k_1<\ldots<k_m\leq n$ of length $m$, then | ||
\[ | \[ | ||
− | \det C=\sum_\beta\det A_{\alpha\beta}\det B_{\beta\alpha}, | + | \det C=\sum_\beta\det A_{\alpha\,\beta}\det B_{\beta\,\alpha}, |
\] | \] | ||
− | where $A_{\alpha\beta}=(a_{\alpha_i\beta_j})$ and $B_{\beta\alpha}=(a_{\beta_j\alpha_i})$. | + | 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. | In case $m>n$, no such $\beta$ exists and the right-hand side above is set to be $0$ by definition. | ||
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\mathrm{rank}C\leq\min\{\mathrm{rank}A,\mathrm{rank}B\}. | \mathrm{rank}C\leq\min\{\mathrm{rank}A,\mathrm{rank}B\}. | ||
\] | \] | ||
+ | Moreover, if $m=2$ and | ||
+ | $$A=\begin{pmatrix} | ||
+ | a_{1}&\dots&a_{n}\\ | ||
+ | b_{1}&\dots&b_{n}\\ | ||
+ | \end{pmatrix} | ||
+ | \quad\text{and}\quad | ||
+ | B=\begin{pmatrix} | ||
+ | a_{1}&b_{1}\\ | ||
+ | \dots&\dots\\ | ||
+ | a_{n}&b_{n}\\ | ||
+ | \end{pmatrix} | ||
+ | $$ | ||
+ | then by Cauchy-Binet | ||
+ | \[ | ||
+ | \sum_{1\leq i<j\leq n}\begin{pmatrix} | ||
+ | a_{i}&a_{j}\\ | ||
+ | b_{i}&b_{j}\\ | ||
+ | \end{pmatrix}^2= | ||
+ | \begin{pmatrix} | ||
+ | \|a\|^2&a\cdot b\\ | ||
+ | a\cdot b&\|b\|^2\\ | ||
+ | \end{pmatrix} | ||
+ | \] | ||
+ | in turn implying Cauchy-Schwartz inequality. |
Revision as of 14:29, 23 November 2012
2020 Mathematics Subject Classification: Primary: 15Axx [MSN][ZBL]
A formula aimed at expressing the determinant of a matrix $C\in M_{m,m}(\mathbb{R})$ that is the product of $A\in\mathrm{M}_{m,n}(\mathbb{R})$ 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 $(k_1,\ldots,k_m)$ with $1\leq k_1<\ldots<k_m\leq n$ of length $m$, then \[ \det C=\sum_\beta\det A_{\alpha\,\beta}\det B_{\beta\,\alpha}, \] 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 \[ \det C=\det A\,\det B. \] A number of interesting consequence of Cauchy-Binet formula is listed below. First of all, an inequality for the rank of the product matrix $C$ follows straightforwardly , i.e., \[ \mathrm{rank}C\leq\min\{\mathrm{rank}A,\mathrm{rank}B\}. \] Moreover, if $m=2$ and $$A=\begin{pmatrix} a_{1}&\dots&a_{n}\\ b_{1}&\dots&b_{n}\\ \end{pmatrix} \quad\text{and}\quad B=\begin{pmatrix} a_{1}&b_{1}\\ \dots&\dots\\ a_{n}&b_{n}\\ \end{pmatrix} $$ then by Cauchy-Binet \[ \sum_{1\leq i<j\leq n}\begin{pmatrix} a_{i}&a_{j}\\ b_{i}&b_{j}\\ \end{pmatrix}^2= \begin{pmatrix} \|a\|^2&a\cdot b\\ a\cdot b&\|b\|^2\\ \end{pmatrix} \] in turn implying Cauchy-Schwartz inequality.
Matteo.focardi/sandbox. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Matteo.focardi/sandbox&oldid=28825