(Using a stride of zero and padding of zero.)
For an input image of size $6\times6$ with $3$ channels (therefore $6\times6\times3$) and a kernel of size $3\times3$ with $3$ channels, the equation for the convolution is given as follows:
$\sum_{m}^{k_1-1}\sum_{n}^{k_2-1}\sum_{C=1}^{3} K_{m,n,c}I_{i+m, j+n,c} + b$
I know that the size of a 2D convolution is given by: $\frac{I-R}{s} + 1 = \frac{6-3}{1}+1 = 4$
So therefore, the first $2$ dimensions will be given as $4\times4$
But what are the dimensions of the 3rd dimension? Is my final dimension going to be $4\times4\times3$ or $4\times4\times1$?