Example 9 For the set $\{1,2, \cdots, n\}$ and each of its non-empty subsets, define a unique “alternating sum” as follows: arrange the numbers in the subset in decreasing order, then alternately subtract and add the successive numbers starting from the largest to get the alternating sum (for example, the alternating sum of $\{1,3,8\}$ is: 8 $-3+1=6)$. For $n=8$, find the total of the “alternating sums” of all subsets.
Solution: Divide all subsets of $\{1,2, \cdots, n\}$, including the set itself and the empty set, into two categories: $M$ and $N$. Place all subsets containing the element $n$ into $M$, and those not containing $n$ into $N$. For a subset $A=\left\{a_{1}, a_{2}, \cdots, a_{k}, n\right\}$ in $M$, there must be a subset $B=\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}$ in $N$; conversely, the same applies. The alternating sum of $A$ is $x=n-a_{k}+a_{k-1}-\cdots$, and the alternating sum of $B$ is $y=a_{k}-a_{k-1}+\cdots$, so $x+y=n$. Therefore, the total alternating sum is $P=n \cdot \frac{2^{n}}{2}=n \cdot 2^{n-1}$. When $n=8$, $P_{8}=8 \times 2^{8-1}=1024$.