Currently I have little background in theory-based linear algebra (I never took an upper-division linear algebra course), so I am asking this question here. Although this is from page 345 of PDE by Evans, 2nd edition, I believe the segment of the textbook's proof I'm about to retype here and my questions to follow concern linear algebra material only.
My questions concerning the proof below are at the bottom of this post.
This proof is for "Weak maximum principle" in Section 6.4 of the textbook. (For the sake of the length of this post, I will not write the theorem here, unless you request me to do so.) I am listing the first two steps of the proof, of which step 2 has the bulk of the linear algebra material. Please focus on step 2 of the proof.
(Note that $U \subset \mathbb{R}^n$.)
Proof. 1. Let us first suppose we have the strict inequality $$Lu < 0 \qquad \text{in $U$,} \tag{5}$$ and yet there exists a point $x_0 \in U$ with $$u(x_0)=\max_{\bar{U}} u. \tag{6}$$ Now at this maximum point $x_0$, we have $$Du(x_0)=0 \tag{7}$$ and $$D^2u(x_0) \le 0. \tag{8}$$
$\quad$2. Since the matrix $A=((a^{ij}(x_0)))$ is symmetric and positive definite, there exists an orthogonal matrix $O=((o_{ij}))$ so that $$OAO^T = \text{diag}(d_1,\ldots,d_n), \quad OO^T=I, \tag{9}$$ with $d_k > 0$ ($k=1,\ldots,n$). Write $y=x_0+O(x-x_0)$. Then $x-x_0 = O^T(y-x_0)$, and so
$$u_{x_i} = \sum_{k=1}^n u_{y_k} o_{ki}, u_{x_i x_j} = \sum_{k,l=1}^n u_{y_k y_l} o_{ki} o_{lj} \quad (i,j=1,\ldots,n).$$ Hence at the point $x_0$, \begin{align} \sum_{i,j=1}^n a^{ij} u_{x_i x_j} &= \sum_{k,l=1}^n \sum_{i,j=1}^n a^{ij} u_{y_k y_l} o_{ki} o_{lj} \\ &= \sum_{k=1}^n d_k u_{y_k y_k} \quad \text{by (9)} \tag{10} \\ &\le 0, \end{align} since $d_k > 0$ and $u_{y_k y_k} (x_0) \le 0$ ($k=1,\ldots,n$), according to $\text{(8)}$.
Am I correct in saying that a positive definite matrix $A$ means $OAO^T = \text{diag}(d_1,\ldots,d_n)$, with $d_k > 0$, and an orthogonal matrix $O = ((o_{ij}))$ gives way to the relation $OO^T = O^T O = I$?
As far as orthogonal matrices are concerned, is $O^{-1}=O^T$, which explains why $y=x_0+O(x-x_0)$ implies $x-x_0=O^T(y-x_0)$?
And finally, my main question was that how do these two consequences (which comprise $\text{(9)}$, by the way) justify this following relation in $\text{(10)}$? $$\sum_{k,l=1}^n \sum_{i,j=1}^n a^{ij} u_{y_k y_l} o_{ki} o_{lj} = \sum_{k=1}^n d_k u_{y_k y_k}$$
Great culinary name, by the way :)
Yes, you're correct on all counts. The equation $OAO^\top = \text{diag}(d_1,\dots,d_n)= D$ can be rewritten $\sum_{i,j} o_{ki}a^{ij}o_{\ell j} = d_k$ when $k=\ell$ and $=0$ otherwise. So $$\sum_{i,j,k,\ell} a^{ij}u_{y_ky_\ell}o_{ki}o_{\ell j} = \sum_{k,\ell}\sum_{i,j} o_{ki}a^{ij}o_{\ell j}u_{y_ky_\ell} = \sum_{k} d_ku_{y_ky_k}.$$