projection onto kernel $A$ and $B$ and $C$ are real $n\times n$ matrices. Define $A$ to be the projection onto the kernel of $C$. Why is the $\operatorname{tr}(B^{t}CA)=0$? My guess was because $CA=0$, and for any vector $x_{n\times 1}$, we have $Ax=y$, where $y\in \ker(C)$, i.e, $Cy=0$.
 A: Your observation is correct. Since $A$ is a projection onto the kernel of $C$, it follows that for every vector $x, A(x)\in\mathrm{ker}(C)$, so $CA(x) = \mathbf{0}$. Therefore, $B^tCA(x)=\mathbf{0}$ for all $x$, hence $B^tCA$ is the zero linear transformation; therefore, the trace is $0$.
Note however, that there is in general no such thing as "the" projection onto a subspace: there are many different projections (unless the subspace is either trivial, in which case the only projection is the zero map, or the whole space, in which case the only projection is the identity). So it's not really correct to talk about "the" projection onto the kernel of $C$.
Instead, one must specify a "direction": we talk about the "projection onto $W$ along $X$" when $W$ and $X$ are subspaces and $V=W\oplus X$. Then the projection maps $w+x$ to $w$. But since there are many possible choices of $X$, there are many possible choices of projections onto $W$.
In the case of $\mathbb{R}^n$, one can simplify matters by using the orthogonal projection onto $W$, but one still must specify that in order to really be able to use the singular definite article "the". 
