Let $\mathbf{D}$ be a diagonal matrix of size $n\times n$ defined over a field $\mathbb{R}$, and $\mathbf{X}$ be a matrix of $n \times n$ for which $\text{rank}(\mathbf{X}) = 1$.
Let $$F_\mathbf{D}(\mathbf{X}) = \sum_{i = 1}^{ n}\sum_{j = 1}^{ n}(d_{ij} -x_{ij})^2$$ Find $\min{F_\mathbf{D}(\mathbf{X})}$ for an arbitrary diagonal matrix $\mathbf{D}$.
Attempt
By Eckart–Young–Mirsky theorem, the best low-rank approximation in terms of Frobenius' norm can be obtained by a singular value decomposition (as if $F_\mathbf{D}{(\mathbf{X} )}$ is in fact a Frobenius' norm $\mid \mathbf{D}-\mathbf{X} \mid$). SVD of a diagonal matrix is the same matrix with its entries sorted in descending order, which is multiplied by permutation matrices.
If this is the case, then the best rank-one approximation is simply the matrix where all the elements except for the largest one are zeroed out. Hence, $\text{min}\space F$ is a sum of squares of $n - 1$ smallest diagonal entries.
Is this correct?
Is there a proof that does not rely on SVD? This concept is outside the scope of the exam program.
Sketch of Proof: For any unit vector $\mathbf u$, we can consider the rank-1 matrices $\mathbf X$ whose column space is spanned by $\mathbf u$. Verify that among these matrices, the matrix that minimizes $F_{\mathbf D}(\mathbf X)$ is $\mathbf u \mathbf u ^T\mathbf D$ (you might want to make use of the fact that the projection $\mathbf u \mathbf u^T \mathbf x$ minimizes $\|\mathbf u - \mathbf x\|^2$). Now, to find the rank-1 matrix $\mathbf X$ that minimizes $F_{\mathbf D}(\mathbf X)$, it suffices to compute $$ \min_{\|\mathbf u\| = 1} F_{\mathbf D}(\mathbf u \mathbf u^T\mathbf D) = \|\mathbf D\|_F^2 - \mathbf u^T \mathbf D^2 \mathbf u. $$ That is, we must maximize $\mathbf u^T \mathbf D^2 \mathbf u$. Verify that if $j = \arg \max_{i} |d_i|$, then this maximum is attained with $\mathbf u = \mathbf e_j$, the $j$th standard basis vector (i.e. the $j$th column of the identity matrix).