Re: [numeric] Convolution requires separable filters

From: Mateusz Loskot (mateusz_at_[hidden])
Date: 2019-07-15 21:11:52

On 19-07-15 16:59:34, stefan wrote:
>On 2019-07-15 4:37 p.m., Mateusz Loskot wrote:
>>The implication is that it requires use of separable kernels [3] (that is
>>kernels with matrix rank equal to 1).
>
>Nit-pick: I think conventional terminology in (multi-)linear algebra
>uses "tensor" to describe objects of arbitrary rank, with "matrix"
>referring to "tensors of rank 2", and vectors as "tensors of rank 1".
>
>There is no such thing as a "matrix of rank 1". At least outside Matlab. :-)

Right, but I meant to write "matrix rank equal to 1" as to mean
"rank of a kernel matrix equals to 1", then such kernel is separable.

>So, 2D convolution filters are separable, and thus can be represented
>as the (outer) product of two vectors.

Yes, but AFAIK not all filters are separable.

Thanks for chiming in with comments, appreciated.

Best regards,
-- 
Mateusz Loskot, http://mateusz.loskot.net
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