Re: [numeric] Convolution requires separable filters

From: stefan (stefan_at_[hidden])
Date: 2019-07-15 20:59:34

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. :-)

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

Stefan

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