Let $\mathcal{V}$ be the invariant subspace of a matrix $A$ corresponding to eigenvalues with negative real parts. How do I show that $P = \frac{1}{2}(I - \operatorname{sign}(A))$ is a projection onto $\mathcal{V}$? Any hint would be appreciated.
Definition: Let $A \in \mathbb{R}^{n\times n}$ with $\Re(\lambda)\neq 0$ for all $\lambda \in \Lambda(A)$. Let
\begin{equation} A = S \begin{bmatrix} J_-&0\\ 0&J_+ \end{bmatrix} S^{-1} \end{equation}
be the Jordan normal form of $A$ where $J_- \in \mathbb{C}^{k \times k}$ and $J_+ \in \mathbb{C}^{n-k \times n-k}$ are such that $\Lambda(J_-) \subseteq \{z \in \mathbb{C}: \Re(z)<0\}$ and $\Lambda(J_+) \subseteq \{z \in \mathbb{C}: \Re(z)>0\}$. Then the sign of $A$ is defined as
\begin{equation} \operatorname{sign}(A) := S \begin{bmatrix} -I&0\\ 0&I \end{bmatrix} S^{-1}. \end{equation}
Let $A \in \mathbb{R}^{n \times n}$ and $P \in \mathbb{R}^{n \times n}$ be matrices. Starting with $P = \frac{1}{2}(I-\operatorname{sign}(A))$ we use the definition of $\operatorname{sign}(A)$ to obtain $P = S \begin{bmatrix} I&0\\ 0&0 \end{bmatrix} S^{-1}$. We use this result to show that there exists a matrix $B\in \mathbb{R}^{n \times n}$ such that $AP = PB$: $\begin{align*} AP &= S \begin{bmatrix} J_-&0\\ 0&J_+ \end{bmatrix} S^{-1} S \begin{bmatrix} I&0\\ 0&0 \end{bmatrix} S^{-1} = S \begin{bmatrix} J_-&0\\ 0&J_+ \end{bmatrix} \begin{bmatrix} I&0\\ 0&0 \end{bmatrix} S^{-1} = S \begin{bmatrix} J_-&0\\ 0&0 \end{bmatrix} \begin{bmatrix} I&0\\ 0&0 \end{bmatrix} S^{-1} \\ &= S \begin{bmatrix} I&0\\ 0&0 \end{bmatrix} \begin{bmatrix} J_-&0\\ 0&0 \end{bmatrix} S^{-1} = S \begin{bmatrix} I&0\\ 0&0 \end{bmatrix} S^{-1} \underbrace{S \begin{bmatrix} J_-&0\\ 0&0 \end{bmatrix} S^{-1}}_{=:B} = PB. \end{align*} $ Let $\mathcal{V} := \operatorname{im}(P)$. Then it follows that $\mathcal{V}$ is $A$-invariant.