Let the SVD of $A \in \mathbb R^{{n}*{n}} $ be given as $A=\sum_{i=0}^n \sigma_{i}u_{i}v_{i}^{T}$ where $\sigma_{1}\gt \sigma_{2}>{...}>\sigma_{n-1}=\sigma_{n}>0 $
Compute a matrix $B$ such that $B$ is the nearest singular matrix to $A$ in $2-norm$ and rank of $(A-B)=3$.
Here by nearest singular matrix we mean that the matrix $B$ will be of rank $n-1$, But how to form such a matrix $B$, such that the rank of $(A-B)=3$?
Please can anyone help me to solve the prblem??
Hint: Suppose, $B = \sum_{i=1}^{n-3}\sigma_{i}u_iv_i^T + 0u_{n-2}v_{n-2}^{T} + 0u_{n-1}v_{n-1}^{T} +0u_{n}v_{n}^{T}$, then $A - B = \sum_{i=n-2}^{n}\sigma_iu_iv_i^{T}. $ Thus, rank of $A-B = 3$.
Useful theorem: Let $k < n$, where $n$ is the rank of $A$ as specified in the question and $A_k =\sum_{i=1}^{k}\sigma_iu_iv_i^T$ then $min_{rank(B)=k}\|A − B\|_2 = \|A − A_k\|_2 = σ_{k+1}$.