How to perform convolution with kernel bigger than image?

Question

In this question I've seen an example of convolution by the kernel with the shape bigger than initial image's one:

import numpy as np
from scipy import signal
x = np.array([(0.51, 0.9, 0.88, 0.84, 0.05), 
              (0.4, 0.62, 0.22, 0.59, 0.1), 
              (0.11, 0.2, 0.74, 0.33, 0.14), 
              (0.47, 0.01, 0.85, 0.7, 0.09),
              (0.76, 0.19, 0.72, 0.17, 0.57)])
y = np.array([(0, 0, 0.0686, 0), 
              (0, 0.0364, 0, 0), 
              (0, 0.0467, 0, 0), 
              (0, 0, 0, -0.0681)])
gradient = signal.convolve2d(np.rot90(np.rot90(y)), x, 'valid')

So, we get this:

array([[ 0.044606, 0.094061], [ 0.011262, 0.068288]])

I understand that y is flipped by 180 degress. But how does "valid" convolution work here? When we can get (2x2) shape from (4x4) convolved by (5x5)?

Answer 1

Is your question how can we perform convolution operation with kernel bigger than input image?

If yes then answer is padding. TL;DR you increase the size of the original image with boundary pixels so that kernel can "fit"