Prove that $1 \leq \|A^{-1}\| _2\|A-B\|_2$

Question

I'm stuck on a numerical analysis problem where I need to prove the following inequality

$$1 \leq \|A^{-1}\| _2\|A-B\|_2,$$

where $A \in \mathbb{R}^{n\times n}$ is regular and $ B \in \mathbb{R}^{n\times n}$ is singular. Given the statement I somehow need to use the singularity of $B$ but I can't find a connection between the 2-norm, the singularity of B and the subtraction of $B$ from $A$.

Answer 1

Let $v$ be a nonzero unit vector in $\ker(B)$. Then $$ \|A^{-1}\|_2\|A-B\|_2 \ge\|I-A^{-1}B\|_2 =\sup_{\|u\|_2=1}\|(I-A^{-1}B)u\|_2 \ge\|(I-A^{-1}B)v\|_2 =\|v\|_2=1. $$

Answer 2

Consider a proof by contrapositive: if $1 > \|A^{-1}\| _2\|A-B\|_2$, then we can conclude that $B$ is non-singular. Indeed, we find that $$ \|I - A^{-1}B\|_2 \leq \|A^{-1}\| _2\|A-B\|_2 < 1, $$ which ensures that the eigenvalues of $I - A^{-1}B$ are all smaller than $1$ in absolute value. Using this, we can conclude that the eigenvalues of $I - (I - A^{-1}B) = A^{-1}B$ are all non-zero, which ensures that $A^{-1}B$ is non-singular. It follows that $B$ is non-singular, which was what we wanted.

As an alternative to the consideration of eigenvalues, you can use the fact $\|I - A^{-1}B\|_2 < 1$ to show that the infinite series $\sum_{k=0}^\infty (I - A^{-1}B)^k$ (known as the Neumann series) converges. Because $\sum_{k = 0}^\infty X^k$ is equal to $(I - X)^{-1}$ if it converges, we can conclude that the series is equal to $[I - (I - A^{-1}B)]^{-1} = (A^{-1}B)^{-1}$, which once again lets us conclude that $A^{-1}B$ is non-singular.