Category Archives: Linear Algebra

Axler’s Linear Algebra Done Right, Exercise 1C-24

Problem
A function f:\mathbb{R} \rightarrow \mathbb{R} is called even if

    \[   f(-x) = f(x)\]


for all x \in \mathbb{R}. A function f:\mathbb{R} \rightarrow \mathbb{R} is called odd if

    \[   f(-x) = -f(x)\]


for all x \in \mathbb{R}. Let U_e denote the set of real-valued even functions on \mathbb{R} and let U_o denote the set of real-valued odd functions on \mathbb{R}. Show that:

    \[   \mathbb{R}^\mathbb{R}=U_e \oplus U_o\]



Proof. To show that U_e + U_o is a direct sum, it is sufficient to prove that U_e \cap U_o = {0}. Let f \in U_e \cap U_o. Hence, f(x) = f(-x) = -f(-x) for all x \in R, which only happens for f(x) = 0. Hence, U_e + U_o is a direct sum.

Let g \in \mathbb{R}^\mathbb{R}. The equation \mathbb{R}^\mathbb{R}=U_e \oplus U_o means that g can be written as the sum of an odd function o and an even function e. Or in other words, for all x \in \mathbb{R}:

(1)   \begin{equation*}    g(x) = o(x) + e(x)\end{equation*}

and also:

(2)   \begin{equation*}    g(-x) = o(-x) + e(-x)\end{equation*}


By the definition of the odd and even functions, (2) can be written as:

(3)   \begin{equation*} g(-x) = -o(x) + e(x)\end{equation*}


The system of linear equations (1) and (3) has a unique solution o(x) = (g(x) - g(-x)) / 2 and e(x) = (g(x) + g(-x)) / 2. Hence, g can always be uniquely written as the sum of an odd function and an even function. \qed