How does one get $u_{y_ky_k}(x_0)\leq 0$ in the proof of the weak maximum principle? The following is the weak maximum principle in Evans's Partial Differential Equations:

Assume $u\in C^2(U)\cap C(\overline{U})$ and $c\equiv0$ in $U$. If $Lu\leq 0$ in $U$, then $\max_{\overline{U}}u=\max_{\partial U}u$. Here
  

Here is the proof by contradiction. Step 1 is an observation in multivariable calculus:



Then an argument in linear algebra is given as the following





Here is my question:
In the very last line, how does one get $u_{y_ky_k}(x_0)\leq 0$? I can see that $u_{x_kx_k}(x_0)\leq 0$ from (8). Why does change of variables not change the ineqaulity? 
 A: This is true, since your function also has the maximum point at $x_0$ in the transformed domain, so (8) still works for the new domain and $u_{y_ky_k}\leq 0$.
A: There is an alternative way to write this down that is sometimes easier to follow. Just note that
$$-\sum_{i,j=1}^n a^{ij}u_{x_ix_j} = -\text{Trace}(AD^2u),$$
where $A = (a^{ij})_{ij}$. The ellipticity condition states that $A$ is positive semi-definite. At a maximum, $D^2u$ is negative semidefinite. Then you just need to show that the multipliction of a real symmetric negative definite matrix with a positive definite matrix is negative definite (and the trace of a negative definite matrix is negative). The calculation is similar to what is in Evans book, but this is perhaps a more "linear algebra" way of looking at it. 
A: This is a companion of Jeff's nice answer. First note that
$$
\frac{\partial x_i}{\partial y_k}=o_{ki}.
$$
By chain rule, 
$$
u_{y_k}=\sum_iu_{x_i}o_{ki},\quad u_{y_ky_k}=\sum_{i,j}u_{x_i,x_j}o_{ki}o_{kj}.
$$
On the other hand, $\sum_{i,j}u_{x_i,x_j}o_{ki}o_{kj}$ is the $kk$-entry of the matrix $O[D^2u]O^T$. 
