Scaling Property in DFT In below image, we have scaling property of DFT, how the final equation is obtained from the above equation. That is how we are getting the scaling factor , $ \frac{1}{|ab|} $ in the final equation ? Please explain how we are getting $ \frac{1}{|ab|} $. Thank you



Equation 4.12 is given as
$\sum\limits_{m=0}^{M-1} \sum\limits_{n=0}^{N-1} f(am,bn)e^{-j \cdot 2\pi(\frac {k}{M} \cdot m + \frac {l}{N} \cdot n)} $  (let $m'=am, n'=bn$)

$=\sum\limits_{m'=0}^{(M-1)a} \sum\limits_{n'=0}^{(N-1)b} f(m',n')e^{-j \cdot 2\pi(\frac {k/a}{M} \cdot m' + \frac {l/b}{N} \cdot n')}$  

($m'$, $n'$ with the incremental interval of $a$ and $b$ respectively)

Note that 
$\sum\limits_{m'=0}^{M-1} \sum\limits_{n'=0}^{N-1} f(m',n')e^{-j \cdot 2\pi(\frac {k/a}{M} \cdot m' + \frac {l/b}{N} \cdot n')}$
$=F(k/a,l/b)$

Here $m'$, $n'$ with the incremental interval of $1$, and the points $m$ and $n$ are evenly distributed on the circle, according to the symmetry, you have 

$\sum\limits_{m'=0}^{(M-1)a} \sum\limits_{n'=0}^{(N-1)b} f(m',n')e^{-j \cdot 2\pi(\frac {k/a}{M} \cdot m' + \frac {l/b}{N} \cdot n')} = \frac {1}{|ab|}F(k/a,l/b)$