So we have the function $f(x)=\frac{1}{2}x^TAx+x^Tb+a$. Here $x,b \in \mathbf{R}^d$. $a \in \mathbf{R}$, $A$ is $d\times d$ symmetric, positive definite matrix with distinct positive eigenvalues. I need to find the stationary points of this function, let $x=\langle x_1,...,x_d\rangle, b= \langle b_1,...,b_d\rangle$.
Ok so $A=UDU^T$ where $U=[u_1,...,u_d]$. $D=diag\{\lambda_1,...,\lambda_d\}$ with $\lambda_1>...>\lambda_d$. Hence, $f(x)=x^TUDU^Tx+x^Tb+a$. $\triangledown f=?$
Your approach is a good one if you switch from $x$-space to $y$-space, $y:=U^Tx$, before searching for the stationary point. Here's another approach. Since $\partial_if=A_{ij}x_j+b_i$, $\nabla f=Ax+b$, so a stationary point satisfies $x=-A^{-1}b$. This point is a minimum, because $\partial_i\partial_jf=A_{ij}$ is positive-definite. In particular, the non-negative function $\frac12(x+A^{-1}b)^TA(x+A^{-1}b)=\frac12x^TAx+b^Tx+\frac12b^TA^{-2}b$ is, up to an additive constant, $f$.
