Theorem: Let $(V, \langle\,,\rangle)$ be a complex inner product space and $T: V \to V$ be a normal operator. Then for any eigenvalue $\lambda \in \textrm{spec}(T)$, there exists an orthogonal projection $P_{\lambda}$ with $\sum_{\lambda \in \textrm{spec}(T)}P_{\lambda}=I$ and $T = \sum_{\lambda \in \textrm{spec}(T)} \lambda P_{\lambda} $.
Proof: Let $\textrm{spec}(T)=\{\lambda_1, ..., \lambda_k\}.$ It has been proved that $V = \bigoplus_{i=1}^kV_{\lambda_i}$, where $V_{\lambda_i}$ is the eigenspace corresponding to eigenvalue $\lambda_i$, $i=1, ..., k$. This implies $\sum_{\lambda \in \textrm{spec}(T)}P_{\lambda}=I$. Applying $T$ to this relation gives the second relation.
Question: How does the decomposition of $V$ into eigenspaces imply the first relation? I can understand how the second is obtained. Could someone help me understand how the decomposition of $V$ into eigenspaces implies the first relation? Thank you for your help.
Suppose $v \in V$ is any vector. Since $V$ is a direct sum of subspaces, namely, $V = \bigoplus_{i=1}^kV_{\lambda_i}$, there exists a unique decomposition of $v$ as a sum of vectors in those subspaces, i.e. there exist unique $v_1 \in V_{\lambda_1}, \dots, v_k \in V_{\lambda_k}$ s.t. $v = v_1 + \dots + v_k$. One important fact you missed in your post is that if $T$ is normal, then the subspaces $V_{\lambda_1}, \dots, V_{\lambda_k}$ are pairwise orthogonal (this fact has probably been given somewhere in your textbook). This implies that for each $i \in \{1, \dots, k\}$, $P_{\lambda_i} v = P_{\lambda_i} (v_1 + \dots + v_k) = v_i$. Therefore, $\left(\sum_{\lambda \in \operatorname{spec}(T)} P_{\lambda}\right)v = v_1 + \dots + v_k = v$. Since $v$ was arbitrary, we get that $\sum_{\lambda \in \operatorname{spec}(T)} P_{\lambda} = I$.