Let $k\in\mathbb{N}, n\in\mathbb{N}, p\in\mathbb{N}, A\in\mathbb{R}^{k\times k}, B\in\mathbb{R}^{k\times n},$ and $C\in\mathbb{R}^{n\times n}$. If $A^{\sf T}+A=0_{k\times k}, B=0_{k\times n},$ and $C^{\sf T}+C=0_{n\times n}$, show that $x^{\sf T}Ax+x^{\sf T}By+y^{\sf T}Cy=0$ for all $x\in\mathbb{R}^k$ and all $y\in\mathbb{R}^n$.
Here is my attempt:
If $A^{\sf T}+A=0_{k\times k}$, then $x^{\sf T}(A^{\sf T}+A)x=0 \ \forall x\in\mathbb{R}^k.$
If $B=0_{k\times n}$, then $x^{\sf T}By=0 \ \forall x\in\mathbb{R}^k$ and $y\in\mathbb{R}^n.$
If $C^{\sf T}+C=0_{n\times n}$, then $y^{\sf T}(C^{\sf T}+C)y=0 \ \forall y\in\mathbb{R}^n.$
Then I do not know how to go on.
Let me give you the gist of what is happening with $A$.
As $A + A^T = 0$ then $$a_{ij} = -a_{ji}$$ where $a_{ij}$ are the entries of $A$. And the diagonal of A is zero.
This implies that for any $x$ vector of size k, $$x^TAx = x_1*(0*x_1 + a_{12}*x_2+...+a_{1k}x_k) +... + x_k *(a_{k1}x_1+...+0*x_k)=x_1x_2 (a_{12} - a_{12})+...+x_{1}x_k(a_{1k}-a_{1k})+...+x_kx_{k-1}(a_{k,k-1}-a_{k,k-1}) = 0$$
Similarly shown for $y^TCy$, and thus as $B$ is zero, and each part is zero, the whole thing is zero.