Let $A \in \mathbb{R}^{m \times n}$ be a matrix with $\text{rank}(A)=\min(m,n)$. The Eckart–Young–Mirsky theorem for the Frobenius norm states that the best approximation of rank $k<\min(m,n)$ for $A$, denoted $A_k$, is:
$$\arg \min_{A_k} \left\Vert A – A_k \right\Vert_F^2 = \sum_{i=1}^k \sigma_i u_i v_i^T$$
where
$$A= \overset{\max(m,n)}{\underset{i=1}{\sum}} \sigma_i u_i v_i^T$$
is the singular value decomposition of $A$ (The singular vectors $u_i$ and $v_i$ are normalized and their corresponding singular values $\sigma_i$ are sorted in descending order).
Is this solution the unique global minimizer?
Best Answer
The minimizer might not be unique. Consider the case where $k=1$ and the eigenvectors corresponding to the largest singular value are not unique. A simple example of that is a rank one approximation of identity matrix.