Definition: Suppose $A\in\mathbb C^{n\times n}$ has no eigenvalues on the imaginary axis and let $S = $ sgn(A). Where $$\text{sgn}(A) :=X\left[\begin{matrix}-I & 0 \\ 0 & I\end{matrix}\right]X^{-1}.$$ and $$X^{-1}AX = \left[\begin{matrix}N & 0 \\ 0 & P\end{matrix}\right]$$ is the Jordan Canonical form for $A$, with $N$ containing all Jordan blocks corresponding to eigenvalues of $A$ in the left half-plane and $P$ containing all Jordan blocks corresponding to eigenvalues in the right half-plane. Furthermore, the dimensions of $-I$ and $I$ are the same as $N$ and $P$, respectively.
Prove that sgn(cA) = sgn(c)sgn(A) for all nonzero real scalars $c$.
I am not quite sure how to prove this. Any help is great! Thanks!
Hints.
Your formula is sufficient to check for $c=-1$.
Let $J$ be a matrix such that $J^{-1}\operatorname{diag}(N,P)J=\operatorname{diag}(P,N)$.
Let $Y=XJ$. Then we get $$ Y^{-1}(-A)Y=\operatorname{diag}(-P,-N). $$
Moreover, if $N$ is a matrix of size $n$ and $P$ is a matrix of size $p$, then we have $$ J^{-1}\operatorname{diag}(-I_n,I_p)J=\operatorname{diag}(I_p,-I_n)\Rightarrow J\operatorname{diag}(I_p,-I_n)J^{-1}=\operatorname{diag}(-I_n,I_p), $$ where $I_k$ is the identity matrix of size $k$.
Finally, $$ \operatorname{sgn}(-A)=Y\operatorname{diag}(-I_p,I_n)Y^{-1}= -X\operatorname{diag}(-I_n,I_p)X^{-1}=-\operatorname{sgn}(A) $$