Is there any underlying relationship between an even function and a symmetric matrix , as well as an odd function and a skew symmetric matrix? Actually, we know that any real function $f:\Bbb R \longrightarrow \Bbb R$ can be uniquely decomposed into an even and an odd function:
$$f(x) = \frac12 \big[ f(x)+f(-x)\big] + \frac12\big[f(x)-f(-x)\big].$$
Again, any square matrix $A$ of order $n$ can be uniquely decomposed into a symmetric and a skew symmetric matrix, both of order $n$.
Again, look into the definition: $f$ is odd $\iff$ $f(-x)=-f(x)$ and $A$ is skew symmetric $\iff$ $A^t=-A$, hence the question arose .
 A: Perhaps you are not using these terms in your mind, but your question can be framed as follows: is there a general (abstract) setting in which a general proposition can be proved, so that both functions from $\Bbb R$ to $\Bbb R$ and square matrices of order $n$ can be seen as special cases, and the two results you mentioned are the application of the general result to these two cases? In other words, can both results be proved in a single proof, in this general setting?
The answer is YES.
Both sets (real functions and square matrices) are vector spaces over $\Bbb R$, and each has a linear operator $T$ $(f(x) \mapsto f(-x)$ for functions, transposition for matrices$)$ with the property that $T \circ T$ is the identity. 

Definitions:
  The transpose of an object is its image under $T$.
  An object is even if it equals its transpose.
  An object is odd if its negative equals its transpose.

For matrices, "even" and "odd" are known as symmetric and skew-symmetric, respectively.

Theorem: Each object has a decomposition as the sum of an even and an odd object (by exactly the same formula used in both special cases).

