Frobenius norm with Kronecker product for rank-1 solution Let $ Y \in \mathbb{C}^{L \times N} $. I need to find the vector $ x $. 
\begin{equation*}
\min_{x}\left\|Y-(1^T\otimes x)\right\|_F^2.
\end{equation*}
My problem is on decoupling $ x $ from the Kronecker operator. Any hints?
 A: So $1 \otimes x$ is a matrix which contains the vector $x$ in every column. Denoting the $j$'th column of $Y$ by $Y_{:j}$, we thus have 
$$L(x) :=  \lVert Y - 1 \otimes x \rVert_F^2 = \sum_{j=1}^N \lVert Y_{:j} - x\rVert_2^2.  $$
Taking the derivative with respect to $x$ gives
$$ \nabla_x L(x) = -2 \sum_{j=1}^N (Y_{:j} - x) = -2 \sum_{j=1}^N (Y_{:j}) + 2Nx.  $$
Setting this to zero yields
$$ x = \frac{1}{N} \sum_{j=1}^N Y_{:j}. $$
A: You can replace the Kronecker product with the outer product of vectors and write
$$\eqalign{
\def\o{{\tt1}}\def\R{{\mathbb R}} 
\min_x\,\left\|Y-x\o^T\right\|^2_F \qquad x\in\R^L\quad \o\in\R^N
}$$
However, this is simply the least squares solution of a linear equation
$$\eqalign{
x\o^T &= Y \\
}$$
whose closed-form solution can be written using the pseudoinverse
$$\eqalign{
x &= Y\,\Big(\o^T\Big)^+
 \;=\; Y\left(\frac{\o}{\o^T\o}\right)
 \;=\; \frac{Y\o}{N} \\
}$$
Derivatives are not needed.
A: Denote $\mathbf{y}_n$ the $n$-th column of $\mathbf{y}$.
The cost function to be minimized can also be written
$$
\phi(\mathbf{x}) = \sum_n \delta_E^2(\mathbf{x},\mathbf{y}_n)
$$
This is similar to compute a Karcher mean wiki using Euclidean distance.
The minimum is reached at the arithmetic mean of the $N$ vectors $\{\mathbf{y}_n\}$.
