Even/odd decomposition is unique: {2,3,4,5,6}

Decomposition exists for any signal

Let any signal be written as the sum of an even part and an odd part:

$$x(t) = x_e(t) + x_o(t) \qquad (1)$$

where $x_e(-t) = x_e(t)$ and $x_o(-t) = -x_o(t)$. Replace $t$ by $-t$:

$$x(-t) = x_e(-t) + x_o(-t) = x_e(t) - x_o(t) \qquad (2)$$

Adding (1) and (2), and subtracting (2) from (1):

$$\begin{aligned} x_e(t) &= \frac{1}{2}\left[x(t) + x(-t)\right] \\ x_o(t) &= \frac{1}{2}\left[x(t) - x(-t)\right] \end{aligned}$$

These parts always exist for any $x(t)$, so every signal can be decomposed into even and odd components. The same holds for DT signals: $x_e[n] = \frac{1}{2}(x[n]+x[-n])$, $x_o[n] = \frac{1}{2}(x[n]-x[-n])$.

Uniqueness

Suppose there are two decompositions: $x = x_{e1} + x_{o1} = x_{e2} + x_{o2}$. Then

$$x_{e1}[n] - x_{e2}[n] = x_{o2}[n] - x_{o1}[n] = g[n]$$

The left side is even, the right side is odd, so $g[n]$ is both even and odd: $g[-n] = g[n]$ and $g[-n] = -g[n]$, which gives $g[n] = 0$ for all $n$. Therefore $x_{e1} = x_{e2}$ and $x_{o1} = x_{o2}$: the decomposition is unique.

Illustration: $x[n] = \{2, 3, 4, 5, 6\}$

No origin is marked, so take the first sample at $n = 0$: $x[0]=2, x[1]=3, x[2]=4, x[3]=5, x[4]=6$, and zero elsewhere. Then $x[-n]$ is the folded sequence on $n = -4 \ldots 0$.

$$\begin{aligned} x_e[n] &= \{3, 2.5, 2, 1.5, \underline{2}, 1.5, 2, 2.5, 3\} \\ x_o[n] &= \{-3, -2.5, -2, -1.5, \underline{0}, 1.5, 2, 2.5, 3\} \end{aligned}$$

Check: $x_e[n] + x_o[n]$ gives $0$ for $n<0$ and $2, 3, 4, 5, 6$ for $n = 0 \ldots 4$, i.e. the original $x[n]$. $x_e$ is symmetric and $x_o$ is antisymmetric with $x_o[0] = 0$. Any other choice of even part would make the odd part non-odd, which confirms uniqueness. (If the origin is placed at a different sample, the parts change, but for that origin they are again unique.)