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 $\lambda_1&gt;...&gt;\lambda_d$. Hence, $f(x)=x^TUDU^Tx+x^Tb+a$. $\triangledown f=?$
It's stated that the gradient of:  $$\frac{1}{2}x^TAx - b^Tx +c$$  is   $$\frac{1}{2}A^Tx + \frac{1}{2}Ax - b$$  How do you grind out this equation? Or specifically, how do you get from $x^TAx$ to $A^Tx + Ax$?
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