Let $A$ be $k\times k$ positive symmetric matrix, $K$ is $k\times d$ full rank matrix with $d<k$, and $v\in\mathbb{R}^k$. I'd like to find $x\in \mathbb{R}^d$ such that $(Kx-v)^TA(Kx-v)$ minimum.
To do this, let $g(x)=(Kx-v)^TA(Kx-v)$. By letting $g'(x)=0 $ yields \begin{equation} K^TAKx-K^TAv+x^TK^TAK-v^TAK=0. \end{equation}
But now, how can we find $x$ satisfying the above equation? Thanks for any help
Expanding $(Kx-v)^TA(Kx-v)$ gives
$$(x^TK^T-v^T)(AKx-Av)=x^TK^TAKx-x^TK^TAv-v^TAKx+v^TAv$$
Since $x^TK^TAv=v^TAKx$ you get the expression
$$x^TK^TAKx-2x^TK^TAv+v^TAv$$
Computing the derivative w.r.t. x and setting it to zero gives
$$2K^TAKx-2K^TAv=0$$
which results in the solution
$$x=(K^TAK)^{-1}K^TAv$$