I would like to prove that convolution of an image $I \in \mathcal{M}_{m_1 \times n_1}$ with respect to a separable 2D filter $F$, (i.e., $F = F_1 F_2$, where $F \in \mathcal{M}_{m_2\times n_2}(\mathbb{R})$, $F_1 \in \mathcal{M}_{m_2\times1}$, and $F_2 \in \mathcal{M}_{1\times n_2}$) is equivalent to convolution with respect to $F_1$ and then $F_2$. If $A*B$ represents convolution of an image matrix $A$ under a filter matrix $B$, then we need to show that $I*F=(I*F_1)*F_2$.
We define convolution as such: $(I*F)_{(m,n)} = \sum_{i=-m_2/2}^{m_2/2}\sum_{j=-n_2/2}^{n_2/2}F[i,j]\cdot I[m-i,n-j]$, or $\sum_{i=-m_2/2}^{m_2/2}\sum_{j=-n_2/2}^{n_2/2} I[i,j]\cdot F[m-i,n-j]$.
Here's what I've tried starting with the left side (using the matrix indexing notation [m,n]):
$((I*F_1)*{F_2}){[m,n]} = \sum_{j=-n_2/2}^{n_2/2}(\sum_{i=-m_2/2}^{m_2/2}{I{[m-i,n-1]}} \cdot {{F_1}{[i,1]}}){[m-1,n-j]}{F_2}{[1,j]}$
Now, I notice that the terms ${F_1}{[i,1]}$ and $F_2[1,j]$ as placed in the summations can be combined to yield $F[i,j]$, however, we are indexing this entire term, so I don't believe we can combine these two terms as such. Additionally, the I don't see how to get terms involving $I$ to become $I[m-i,n-j]$.
Any advice would be much appreciated.