Find norm of image convolution
$begingroup$
Suppose there is matrix A that is a black and white image. There's also matrix B - a filter.
SVD decomposition of matrix B is also given
$$ B = USV^T$$
Let's check the first main component of SVD decomposition: $B_0$
$$B = sum_{i = 0}^{p}B_i$$
$$B_i = u^{(i)}sigma_i(v^{(i)})^T$$, where $B_i$ - $i$-th main component.
I have to evaluate: $$||A*B - A*B_0||$$
Where * is the convolution operator of image with filter.
$$(A * B)_{ij} = sum_{p, q}a_{pq}b_{i - p, j - q}$$
I know that
$$||B - B_0||_2 = ||U(Sigma-Sigma_0)V^T||_2 = ||diag(0,sigma_{1}, ... ,sigma_p)||_2 = sigma_{max} = sigma_{1}$$
But don't know how to evaluate
$||A*B - A*B_0||$ ?
I tried:
$||A*B - A*B_0|| = ||A * (B-B_0)||$ but what do I do then?
matrices norm convolution
asked Dec 6 '18 at 19:21
Gusev SlavaGusev Slava
1155
$endgroup$
add a comment |
$begingroup$
Suppose there is matrix A that is a black and white image. There's also matrix B - a filter.
SVD decomposition of matrix B is also given
$$ B = USV^T$$
Let's check the first main component of SVD decomposition: $B_0$
$$B = sum_{i = 0}^{p}B_i$$
$$B_i = u^{(i)}sigma_i(v^{(i)})^T$$, where $B_i$ - $i$-th main component.
I have to evaluate: $$||A*B - A*B_0||$$
Where * is the convolution operator of image with filter.
$$(A * B)_{ij} = sum_{p, q}a_{pq}b_{i - p, j - q}$$
I know that
$$||B - B_0||_2 = ||U(Sigma-Sigma_0)V^T||_2 = ||diag(0,sigma_{1}, ... ,sigma_p)||_2 = sigma_{max} = sigma_{1}$$
But don't know how to evaluate
$||A*B - A*B_0||$ ?
I tried:
$||A*B - A*B_0|| = ||A * (B-B_0)||$ but what do I do then?
matrices norm convolution
asked Dec 6 '18 at 19:21
Gusev SlavaGusev Slava
1155
$endgroup$
add a comment |
$begingroup$
Suppose there is matrix A that is a black and white image. There's also matrix B - a filter.
SVD decomposition of matrix B is also given
$$ B = USV^T$$
Let's check the first main component of SVD decomposition: $B_0$
$$B = sum_{i = 0}^{p}B_i$$
$$B_i = u^{(i)}sigma_i(v^{(i)})^T$$, where $B_i$ - $i$-th main component.
I have to evaluate: $$||A*B - A*B_0||$$
Where * is the convolution operator of image with filter.
$$(A * B)_{ij} = sum_{p, q}a_{pq}b_{i - p, j - q}$$
I know that
$$||B - B_0||_2 = ||U(Sigma-Sigma_0)V^T||_2 = ||diag(0,sigma_{1}, ... ,sigma_p)||_2 = sigma_{max} = sigma_{1}$$
But don't know how to evaluate
$||A*B - A*B_0||$ ?
I tried:
$||A*B - A*B_0|| = ||A * (B-B_0)||$ but what do I do then?
matrices norm convolution
asked Dec 6 '18 at 19:21
Gusev SlavaGusev Slava
1155
$endgroup$
Suppose there is matrix A that is a black and white image. There's also matrix B - a filter.
SVD decomposition of matrix B is also given
$$ B = USV^T$$
Let's check the first main component of SVD decomposition: $B_0$
$$B = sum_{i = 0}^{p}B_i$$
$$B_i = u^{(i)}sigma_i(v^{(i)})^T$$, where $B_i$ - $i$-th main component.
I have to evaluate: $$||A*B - A*B_0||$$
Where * is the convolution operator of image with filter.
$$(A * B)_{ij} = sum_{p, q}a_{pq}b_{i - p, j - q}$$
I know that
$$||B - B_0||_2 = ||U(Sigma-Sigma_0)V^T||_2 = ||diag(0,sigma_{1}, ... ,sigma_p)||_2 = sigma_{max} = sigma_{1}$$
But don't know how to evaluate
$||A*B - A*B_0||$ ?
I tried:
$||A*B - A*B_0|| = ||A * (B-B_0)||$ but what do I do then?
matrices norm convolution
matrices norm convolution
asked Dec 6 '18 at 19:21
Gusev SlavaGusev Slava
1155
asked Dec 6 '18 at 19:21
Gusev SlavaGusev Slava
1155
asked Dec 6 '18 at 19:21
Gusev SlavaGusev Slava
1155
asked Dec 6 '18 at 19:21
Gusev SlavaGusev Slava
1155
asked Dec 6 '18 at 19:21
Gusev SlavaGusev Slava
1155
1155
add a comment |
add a comment |
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