Problem: Prove that $\left\{ A \in \mathbb{R}^{n \times n} \mid A \text{ is symmetric}\right\}^{\bot} = \left\{ A \in \mathbb{R}^{n \times n} \mid A \ \text{is skew-symmetric}\right\}$ with $\langle A, B \rangle = Tr(A^T B)$.
Attempt at proof: Let $A$ be symmetric, and $B$ skew-symmetric. I want to prove that $\langle A, B \rangle = 0$. So this is what I had so far: \begin{align*} \langle A, B \rangle &= Tr(A^T B) \\ &= Tr(AB) \\ &= \sum_{i=1}^n (AB)_{ii} \\ &= \sum_{i=1}^n \sum_{k=1}^n (a_{ik} b_{ki}) \end{align*} Now I need to use somewhere the fact that $b_{ii} = 0$, i.e. the diagonal elements of a skew-symmetric matrix are zero. But I don't know how to split up the summations? Help would be appreciated!
You're on the right track. You don't need to break it down into components, assuming you already know the following: $$ \text{Tr}(AB) = \text{Tr}(BA)~,~~\text{Tr}(A^TB) = \text{Tr}(B^TA)~,~~\text{hence}~ \langle A, B\rangle = \langle B, A\rangle $$ I assume you've already proved these things previously, by breaking down into components. So now we can proceed like at the start of your attempt: $$ \langle A,B\rangle = \text{Tr}(A^TB) = \text{Tr}(AB) = \text{Tr}(BA) = \text{Tr}(-B^TA) = \langle -B, A\rangle = - \langle A, B\rangle $$
To show that the spaces of symmetric and anti-symmetric matrices are actually orthogonal complements of each other, we also need to show that any matrix has a unique decomposition as a sum of a symmetric matrix and an anti-symmetric one. But this is easy: $$ A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T) $$ The first term is symmetric; the second is anti-symmetric.