Orthogonal Matrix of Maximum Principle

Question

Let $A = ((a^{ij}(x_{0})))$ be a symmetric and positive definite matrix with the size of $n \times n$. Thus, there exists an orthogonal matrix $O=((o_{ij}))$ such that $OAO^{T}=\text{diag}(d_{1},...,d_{n})$ and $OO^{T}=I$ with $d_{k}>0$ for $k=1,2,3...,n$. Write $y = x_{0} + O(x-x_{0})$, then $x-x_{0}=O^{T}(y-x_{0})$.

There is something I want to confirm since my calculation is not similar with the original source (Partial Differential Equations - L.C. Evans - Theorem 1 section 6.4.1 (Weak Maximum Principle)).

So, this is my calculation:
Given $u(x)\in C^{2}(U)\cap C(\bar{U})$ for an open set $U \subset \mathbb{R}^{n}$, I have the following equations: $$\begin{bmatrix}y_{1}\\y_{2}\\.\\.\\.\\y_{n}\end{bmatrix} = \begin{bmatrix}x_{0_{1}}+\sum_{j=1}^{n}o_{1j}(x_{j}-x_{0_{j}})\\x_{0_{2}}+\sum_{j=1}^{n}o_{2j}(x_{j}-x_{0_{j}})\\.\\.\\.\\x_{0_{n}}+\sum_{j=1}^{n}o_{nj}(x_{j}-x_{0_{j}})\end{bmatrix}$$ Therefore, for any $k\in{1,2,3,...,n}$, we will have
$$y_{k}=x_{0_{k}}+\sum_{j=1}^{n}o_{kj}(x_{j}-x_{0_{j}})$$ $$\frac{\partial y_{k}}{\partial x_{i}} = o_{ki}$$ Now, for $u_{x_{i}}= \sum_{k=1}^{n}u_{y_{k}}\frac{\partial y_{k}}{\partial x_{i}}=\sum_{k=1}^{n}u_{y_{k}}o_{ki}$

But, the book instead states that $u_{x_{i}}= \sum_{k=1}^{n}u_{y_{k}}o_{ik}$. Where do I make mistakes on my calculation?