Conditions for $(I-x A)$ to be contractive for some $x\in \mathbb{R}$ 
*

*Real parts of eigenvalues of matrix $A$ must have same sign for $I-xA$ to be contractive for some $x\in \mathbb{R}$


*Suppose no 0 eigenvalues. Is it also a sufficient condition?
 A: The conclusion is not true in general. Consider the matrix
$$A=\begin{pmatrix} 1& 2\\
0 & 1
\end{pmatrix}$$
Then $I-xA$ is not contractive for any $x\in \mathbb{R}\setminus \{0\}.$ Indeed
$$I-xA=\begin{pmatrix} 1-x& -2x\\
0 & 1-x
\end{pmatrix}$$
Hence $I-xA$ is not contractive for $x<0$ and $x>{1\over 2}.$ We can restrict to $0<x\le {1\over 2}.$
We have
$$B_x:=(I-xA)^*(I-xA)=\begin{pmatrix}(1-x)^2 & -2x(1-x)\\
-2x(1-x) & 4x^2+(1-x)^2
\end{pmatrix}$$
Then $\det B_x=(1-x)^4$ and ${\rm tr}\, B_x=2(1-x)^2+4x^2.$ The characteristic equation is
$$\lambda^2-2[(1-x)^2+2x^2]\,\lambda +(1-x)^4=0$$
One of the roots is equal
$$\displaylines{\lambda=1 - 2 x + 3 x^2 + 2x\sqrt{1 - 2 x + 2 x^2}\\ \ge 1-2x+3x^2+2x(1-x)=1+x^2>1}$$
Thus $B_x$ is not contractive, so neither is $I-xA.$
The conclusion is true if $A$ is a normal matrix. In that case $A$ is unitarily equivalent to the diagonal matrix $D$ with the eigenvalues of $A$ on the main diagonal. We have $\|I-xA\|=\|I-xD\|.$ The matrix $I-xD$ is contractive if and only if $|1-x\lambda|\le 1$ for every eigenvalue $\lambda$ of $A.$ Assume the real part of $\lambda$ is positive for every $\lambda.$ Hence $\lambda=u+iv,$ where $u>0.$
Then $$|1-x\lambda|^2=(1-xu)^2+x^2v^2=1-2ux+(u^2+v^2)x^2$$
Thus $|1-x\lambda|<1$ for $0<x<2u/(u^2+v^2).$ Therefore $x$ should satisfy $$0<x <\min_{\lambda} {2{\rm Re }\,\lambda\over |\lambda|^2}$$
Similar analysis can be made if $u<0$ for all eigenvalues $\lambda.$
