Given vectors $\boldsymbol{a}_k$ for $k=1,\dots,n$, and vector $\boldsymbol{x}$, all in $\mathbb{R}^d$. Does there exist a closed form solution for the following problem? \begin{equation} \boldsymbol{x}^* = \arg\min_{\boldsymbol{x}} \sum_k \| \frac{\boldsymbol{a}_k^T \boldsymbol{x}}{\|\boldsymbol{x}\|^2} \boldsymbol{x} -\boldsymbol{a}_k \|^2 \,. \end{equation}
The trivial case is when $n=1$, in which case obviously $\boldsymbol{x}^*=c\, \boldsymbol{a}_1$, where $c$ is any non-zero scalar. However, I am curious if closed form exists also for when $n>1$.
Thanks
Golabi
Let $A$ be the matrix with the $a_k$ as rows. Then \begin{align*} \sum_k\left\| \frac{\boldsymbol{a}_k^T \boldsymbol{x}}{\|\boldsymbol{x}\|^2} \boldsymbol{x} -\boldsymbol{a}_k \right\|^2 &= \sum_k\left(\frac{|a_k^Tx|^2}{\|x\|^4}\|x\|^2 - 2\frac{a_k^Tx}{\|x\|^2}x^Ta_k + \|a_k\|^2 \right)\\ &= \sum_k\|a_k\|^2 - \sum_k\frac{|a_k^Tx|^2}{\|x\|^2}\\ &= Tr(A^TA) - \sum_k\left|a_k^T\frac x{\|x\|}\right|^2\\ &= Tr(A^TA) - \left\|A\frac x{\|x\|}\right\|^2. \end{align*} Hence, your function of $x$ is independent of the norm of $x$. Thus, it suffices to minimize over $\{x : \|x\|=1\}$ and the minimum equals $Tr(A^TA) - \|A\|^2$, where $\|A\|$ is the operator norm of $A$. The minimum is thus equal to $Tr(A^TA)-\|A^TA\|$ and $\|A^TA\|$ is the largest eigenvalue of $A^TA$. The maximum of $\|Ax\|^2$ is therefore attained at each normalized eigenvector $x$ of $A^TA$ corresponding to the largest eigenvalue of $A^TA$.