Singular Matrix and Identity Matrix

Question

Let $I$ be an identity matrix.

If $I+A$ is a singular matrix (i.e. $I+A$ has at least one eigenvalue being zero), then $\|A\|\geq 1$.

How does a zero eigenvalue of $I+A$ imply $\|A\|\geq 1$?

Answer 1

If $I+A$ is singular, then $(I+A)x=0$ for some vector $x$. In other words, $Ax=-x$, and hence $-1$ is an eigenvalue of $A$. This implies $\rho(A) \geq 1$. Since $\Vert A \Vert \geq \rho(A)$ for any norm, the result follows.

Answer 2

The spectral mapping theorem gives $$\sigma(I + A) = 1 + \sigma(A) = \{1+\lambda : \lambda \in \sigma(A)\}$$ so if $0 \in \sigma(I + A)$ then $-1 \in \sigma(A)$.

Hence there exists $x \ne 0$ such that $Ax = -x$.

We conclude

$$\|A\| \ge \frac{\|Ax\|}{\|x\|} = \frac{\|-x\|}{\|x\|} = 1$$