Corollary

Let $\lambda \in$ Sp($u$). The image of $p_{\{\lambda\}}$ is the proper subspace of $u$ relative to $\lambda$.

Let $x\in$ dom($u$). Let $\mu$ (resp. $\nu$ ) denote the spectral measure of the element $x$ relative to $u$ (resp. the spectral measure of $(x, u(x))$ relative to $u$). One has $\nu =$ Id$_{Sp(u)}\cdot \mu$ (Corollary to Proposition 5 of IV, p. 275). If $u(x) =\lambda x$, one also has $\nu =\lambda \mu$, whence the equality Id$_{Sp(u)}\cdot \mu=\lambda \mu$. This implies that the support of $\mu$ is contained in $\{\lambda \}($cf. INT, V, p. 46, § 5, n$^o3$, Corollary 2) and therefore $x$ belongs to the image of $p_{\{\lambda\}}$ (Proposition 9, a)).

Conversely, suppose that $x$ belongs to the image $E_{\lambda}$ of $p_{\{\lambda\}}$. Then $x$ belongs to dom($u$) and $u(x)$ also belongs to $E_{\lambda}($loc. cit., b)). Since $\varphi_{\{\lambda\}}\cdot$ (Id$_{Sp(u)}-\lambda$ ) $= 0$, one has the relation $p_{\{\lambda\}}\circ (u-\lambda 1_E)\subset 0$ (Proposition 5 of IV, p. 275, c)), whence $0 =p_{\{\lambda\}}(u(x))-\lambda p_{\{\lambda\}}(x) =u(x)-\lambda x$.