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    Even and odd function subspaces sum to $V$ with zero intersection

    E-HK-22-8

    Area
    Algebra
    Topics
    Linear Algebra, Subspaces
    Status
    draft

    Let \(V\) be the vector space of all functions from \(\RR\) into \(\RR\); let \(V_e\) be the subset of even functions, \(f(-x) = f(x)\); let \(V_o\) be the subset of odd functions, \(f(-x) = -f(x)\).

    1. Prove that \(V_e\) and \(V_o\) are subspaces of \(V\).

    2. Prove that \(V_e + V_o = V\).

    3. Prove that \(V_e \cap V_o = \{0\}\).

    Solution

    1. \(V_e\) and \(V_o\) are subspaces of \(V\).

    The zero function is both even and odd. If \(f,g\) are even and \(a,b\in\RR\), then \[ (af+bg)(-x)=af(-x)+bg(-x)=af(x)+bg(x), \] so \(af+bg\) is even. If \(f,g\) are odd, then \[ (af+bg)(-x)=-af(x)-bg(x)=-(af+bg)(x), \] so \(af+bg\) is odd.

    2. \(V_e+V_o=V\).

    For \(f\in V\), define \[ f_e(x)=\frac{f(x)+f(-x)}2, \qquad f_o(x)=\frac{f(x)-f(-x)}2. \] Then \(f_e\) is even, \(f_o\) is odd, and \[ f=f_e+f_o. \] Thus every function belongs to \(V_e+V_o\).

    3. \(V_e\cap V_o=\{0\}\).

    If \(f\) is both even and odd, then for every \(x\), \[ f(x)=f(-x)=-f(x). \] Hence \(2f(x)=0\), so over \(\RR\) one has \(f(x)=0\) for all \(x\). Thus \(f=0\).

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