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$A$ 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, $Cy=0$.

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@Arturo Magidin: the only thing mentioned in the problem's solution is that $A$ is a projection onto the kernel of $A$. The TA might mean something in his head and he didn't write it explicitely.So, his solution doesn't specify which projection. For me, I saw it like this: If I consider any $x_{n\times 1}$, then $Ax=y$, where $y\in Ker(C)\Rightarrow Cy=0$. So, it follows that $CAx=Cy=0$. So, $CAx=0$, for any $x$, implies that $CA=0.I$. Does that make sense? If this depends on the type of projection, can you please tell me what happens in each type? –  mchris619 Apr 29 '12 at 21:07
    
@Arturo Magidin. I have checked the problem's solution again. It should be $tr(B^{t}CA)$ instead of $tr(B^{t}AC)$. Sorry, my bad! I will edit the original statement. Thanks a lot for clarifying this... –  mchris619 Apr 29 '12 at 21:20
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Ah, well; then, absolutely! If $A$ projects onto the kernel of $C$, in any which way, the $\mathrm{Im}(A)\subseteq \mathrm{ker}(C)$, so $CA=0$, hence $B^tCA = 0$, so the trace is $0$. After you fix the statement, I'll delete my comments, and you can post your solution as an answer! Just note that in your final sentence, you may want "hence" instead of "i.e." (which means "that is"). –  Arturo Magidin Apr 29 '12 at 21:26
    
@Arturo Magidin: I will definitely post my solution for this question later on. Can I add some of the details you provided in the comment which is above this to my answer? –  mchris619 Apr 29 '12 at 21:28
    
Of course you can. –  Arturo Magidin Apr 29 '12 at 21:32
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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".

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