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Is the product of two “sample” matrices a sample matrix?
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$begingroup$
Let $f,g colon mathbb R^2 to mathbb R$ be two smooth functions.
Take a uniform, square grid of the unit cube in $mathbb R^2$ and let ${p_{ij}}_{i,j=1, ldots, L} subset mathbb R^2$ be the points of the grid. Let $f_{i,j} = f(p_{i,j})$ and $g_{i,j} = g(p_{i,j})$. Consider the matrices $A = (f_{ij})_{i,j=1, ldots, L}$ and $B = (g_{ij})_{i,j=1, ldots, L}$.
Finally, let us agree that we call a $L times L$ matrix $C$ a sample matrix if there exists a function $h$ as above such that $c_{ij} = h(p_{ij})$ for every $i,j$. In particular, $A$ is the sample matrix of $f$ and $B$ of $g$.
Q. Is the matrix product $D := AB$ a sample matrix?
The elements are $d_{ij} := sum_l f_{il} g^{lj}$ and this seems a close relative of the convolution $f ast g$ (or of the Riemann-Stiltjes integral of $f$ in $dg$). However the matrix $D$ cannot be the sample matrix of $f ast g$ (because the convolution is a commutative operation, while matrix product is not).
linear-algebra matrices multivariable-calculus approximation numerical-calculus
$endgroup$
add a comment |
$begingroup$
Let $f,g colon mathbb R^2 to mathbb R$ be two smooth functions.
Take a uniform, square grid of the unit cube in $mathbb R^2$ and let ${p_{ij}}_{i,j=1, ldots, L} subset mathbb R^2$ be the points of the grid. Let $f_{i,j} = f(p_{i,j})$ and $g_{i,j} = g(p_{i,j})$. Consider the matrices $A = (f_{ij})_{i,j=1, ldots, L}$ and $B = (g_{ij})_{i,j=1, ldots, L}$.
Finally, let us agree that we call a $L times L$ matrix $C$ a sample matrix if there exists a function $h$ as above such that $c_{ij} = h(p_{ij})$ for every $i,j$. In particular, $A$ is the sample matrix of $f$ and $B$ of $g$.
Q. Is the matrix product $D := AB$ a sample matrix?
The elements are $d_{ij} := sum_l f_{il} g^{lj}$ and this seems a close relative of the convolution $f ast g$ (or of the Riemann-Stiltjes integral of $f$ in $dg$). However the matrix $D$ cannot be the sample matrix of $f ast g$ (because the convolution is a commutative operation, while matrix product is not).
linear-algebra matrices multivariable-calculus approximation numerical-calculus
$endgroup$
add a comment |
$begingroup$
Let $f,g colon mathbb R^2 to mathbb R$ be two smooth functions.
Take a uniform, square grid of the unit cube in $mathbb R^2$ and let ${p_{ij}}_{i,j=1, ldots, L} subset mathbb R^2$ be the points of the grid. Let $f_{i,j} = f(p_{i,j})$ and $g_{i,j} = g(p_{i,j})$. Consider the matrices $A = (f_{ij})_{i,j=1, ldots, L}$ and $B = (g_{ij})_{i,j=1, ldots, L}$.
Finally, let us agree that we call a $L times L$ matrix $C$ a sample matrix if there exists a function $h$ as above such that $c_{ij} = h(p_{ij})$ for every $i,j$. In particular, $A$ is the sample matrix of $f$ and $B$ of $g$.
Q. Is the matrix product $D := AB$ a sample matrix?
The elements are $d_{ij} := sum_l f_{il} g^{lj}$ and this seems a close relative of the convolution $f ast g$ (or of the Riemann-Stiltjes integral of $f$ in $dg$). However the matrix $D$ cannot be the sample matrix of $f ast g$ (because the convolution is a commutative operation, while matrix product is not).
linear-algebra matrices multivariable-calculus approximation numerical-calculus
$endgroup$
Let $f,g colon mathbb R^2 to mathbb R$ be two smooth functions.
Take a uniform, square grid of the unit cube in $mathbb R^2$ and let ${p_{ij}}_{i,j=1, ldots, L} subset mathbb R^2$ be the points of the grid. Let $f_{i,j} = f(p_{i,j})$ and $g_{i,j} = g(p_{i,j})$. Consider the matrices $A = (f_{ij})_{i,j=1, ldots, L}$ and $B = (g_{ij})_{i,j=1, ldots, L}$.
Finally, let us agree that we call a $L times L$ matrix $C$ a sample matrix if there exists a function $h$ as above such that $c_{ij} = h(p_{ij})$ for every $i,j$. In particular, $A$ is the sample matrix of $f$ and $B$ of $g$.
Q. Is the matrix product $D := AB$ a sample matrix?
The elements are $d_{ij} := sum_l f_{il} g^{lj}$ and this seems a close relative of the convolution $f ast g$ (or of the Riemann-Stiltjes integral of $f$ in $dg$). However the matrix $D$ cannot be the sample matrix of $f ast g$ (because the convolution is a commutative operation, while matrix product is not).
linear-algebra matrices multivariable-calculus approximation numerical-calculus
linear-algebra matrices multivariable-calculus approximation numerical-calculus
edited 20 hours ago
Romeo
asked Feb 13 at 18:29
RomeoRomeo
3,05721148
3,05721148
add a comment |
add a comment |
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