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An even function is one in which $f(x)=f(-x)$.

For two variables I believe this is $f(x,y)=f(-x,-y)$

If I wish to make a 2D even matrix how would I do this?

$$ \begin{matrix} (0,0) & (0,1) \\ (1,0) & (1,1) \end{matrix}$$

Looking at the indices I can't see any pattern that would allow a even matrix.

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We are looking for a linear transformation $T : V \to W$ such that $T(u) = T(-u)$, or equivalently that $T(u) - T(-u) = 0$. Since $T$ is linear, we have \begin{align} T(u) - T(-u) &= T(u) + T(u) \\ &= T(u + u) \\ &= 2T(u), \end{align} so we are looking for $T$ such that for all $u \in V$, $2T(u) = 0$, i.e., $$ T(u) = 0. $$ Only one such transformation exists, namely: $T = 0$.

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  • $\begingroup$ So there's only a trivial version in which the matrix is all zeros? $\endgroup$
    – Kevinj22
    May 14, 2018 at 15:22
  • $\begingroup$ Exactly. Incidentally, this is a great example for when using the abstract notions of linear algebra greatly clarifies the specific $\mathbb{R}^2$ case you were considering. $\endgroup$
    – David Kraemer
    May 14, 2018 at 15:24

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