Let $V$ be a real vector space and $E$ an idempotent linear operator on $V$, i.e., a projection. Prove that $(I + E)$ is invertible. Find $(I + E)^{-1}$.
My attempt: Suppose $E:V\to V$ be a projection. Then $V=R_E\oplus N_E$. Let $B_1=\{\alpha_1,…,\alpha_r\}$ and $B_2=\{\alpha_{r+1},…,\alpha_n\}$ be basis of $R_E$ and $N_E$, respectively. So $B=B_1\cup B_2=\{\alpha_1,…,\alpha_n\}$ is basis of $V$ and $[E]_B=\begin{bmatrix}I_r&0 \\ 0&0\\ \end{bmatrix}$, where $I_r\in M_{r\times r}(\Bbb{R})$ is identity and $r\geq 0$. By exercise 6 section 5.4, $(I+E)$ is invertible$\iff$$\det (I+E)\neq 0$. So $$\det (I+E)=\det ([I+E]_B)=\det ([I]_B+[E]_B)=\det \left(I_n+\begin{bmatrix} I_r&0 \\ 0&0\\ \end{bmatrix}\right)=\det \left(\begin{bmatrix} 2I_r&0 \\ 0&I_{n-r}\\ \end{bmatrix}\right)=2^r\gt 0.$$ Thus $(I+E)$ is invertible. By theorem 4 section 5.4, $(I+E)^{-1}=\frac{1}{2^r}\cdot \text{adj}(I+E)$. Is my proof correct?
Best Answer
Since that $E$ is an idempotent linear operator on $V$, we have:
$$E^2-E-2I=-2I$$
i.e. $$(I+E)(2I - E)=2I$$
hence $(I+E)^{-1}$ is equal to $\frac 12 (2I-E)$.