The exercise says: Matrices $A, B\in\mathbb{R^{m\times n}}$ have the same rank $r$ where $r\leq n\leq m$. Prove that $$\sigma_r^A - \sigma_r^B\leq ||A - B||_2\text{,}$$ where $\sigma_i^C$ is the $i$-th singular value of the matrix $C$, where $\sigma_1^C\geq\sigma_2^C\geq \cdots$

I really do not know how to start. My tries:

  • $\sigma_1^A - \sigma_1^B =||A||_2 - ||B||_2\leq ||A-B||_2$, but the difference between the first singular values can be smaller than the difference between $r$-th singular values
  • compute eigenvalues of $(A-B)^T(A-B)$: do knot know how to do that


  • $\begingroup$ Try the Courant-Fischer min-max theorem. $\endgroup$ – Algebraic Pavel Sep 5 '18 at 20:39

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