Relation between perturbed matrix and condition number of the matrix If A is non‐singular but the perturbed matrix (A+δA) is singular, then show that $$∥A∥/∥δA∥≤y $$ Where y is condition number of the matrix A. Tried for a solution The relation $$(A+δA)(x+δx)=b $$ implies that $$Ax+δA(x+δx)+Aδx=b$$ and since $$Ax=b$$ the above becomes $$δA(x+δx)+Aδx=0$$ and hence $$δx=−A^{−1}δA(x+δx)$$ which implies that $$∥δx∥≤∥A^{−1}∥∥δA∥∥x+δx∥=y⋅\frac {∥δA∥}{∥A∥} ∥x+δx∥,$$ and finally $$\frac {\frac{∥δx∥}{∥x+δx∥}} {\frac {∥δA∥}{∥A∥}}≤y$$ $$\frac{∥δx∥}{∥x+δx∥} {\frac {∥A∥}{∥δA∥}}≤y$$ This what I get so far ! I don't know how to remove $$\frac{∥δx∥}{∥x+δx∥}$$

If $A+E$ is singular, $(A+E)x=0$ for some $x\
eq 0$, so $$ x=-A^{-1}Ex\implies\|x\|\leq\|A^{-1}\|\|E\|\|x\|\implies 1\leq\|A^{-1}\|\|E\| \implies\frac{\|A\|}{\|E\|}\leq\|A^{-1}\|\|A\|. $$