# The singular values of the best rank-$k$ approximation to a matrix

Let $$A\in\mathbb{C}^{m\times n}$$ be a complex matrix. Let $$B_k$$ be a best rank-$$k$$ approximation to $$A$$ such that $$\begin{equation*} B_k\in\arg\min\limits_{{\rm rank}(B)=k}||A-B||_F, \end{equation*}$$ where $$||\cdot||_F$$ denotes the Frobenius norm. Then, I guess $$k$$ largest singular values of $$B_k$$ is identical with the $$k$$ largest singular values of $$A$$ and other singular values of $$B_k$$ are all zero. Is my conjecture correct? Can anyone help me prove this?

• By pre- and post- multiplication by unitary matrices, you may as well assume $A$ is diagonal. Then any besta pproximation to $\|A-B_k\|$ will be by diagonal $B_k$, and so on. – kimchi lover May 11 '19 at 7:46
• why the best rank-k approximation to diagonal matrix should be also diagonal? – Lin Xuelei May 11 '19 at 9:52
• The map that replaces off-diagonal entries with zero is an orthogonal projection onto the set of diagonal matrices. So you are left with a combinatorial problem, of selecting which $k$ diagonal elements to preserve and which others to set to 0. – kimchi lover May 11 '19 at 13:31
• the feasible set is $\{B|{\rm rank(B)}=k\}$. A rank-$k$ matrix may have more than $k$ nonzero diagonal entries. So, what's the usage of the orthogonal projection here. could you provide more details? – Lin Xuelei May 11 '19 at 16:04