A normal endomorphism is a sum of its eigenvalues times pairwise orthogonal projections, and each spectral projection is a polynomial in the endomorphism

By [L1], $V$ has an orthonormal eigenbasis. If $V=0$, then there are no eigenvalues and $T=0$ is the empty sum. Otherwise $V=E_{{\lambda }_1}(T)\oplus \cdots \oplus E_{{\lambda }_r}(T)$ with the summands pairwise orthogonal; if $P_j$ is the orthogonal projection onto $E_{{\lambda }_j}(T)$ and $v=v_1+\cdots +v_r$ with $v_j\in E_{{\lambda }_j}(T)$, then $Tv={\lambda }_1{v}_1+\cdots +{\lambda }_r{v}_r={\sum }_{j=1}^r{\lambda }_j{P}_{j}v$, hence $T={\sum }_{j=1}^r{\lambda }_j{P}_j$.

Because the eigenspaces are pairwise orthogonal and the $P_j$ are the corresponding orthogonal projections, one has $P_j^2=P_j$, $P_j^{\ast }=P_j$, and $P_i{P}_j=0$ for $i\mathrm{\ne }j$, so the $P_j$ are pairwise orthogonal projections in the sense of [L2].

Since $T$ is diagonalisable with distinct eigenvalues, its primary decomposition is exactly the direct sum of the eigenspaces $E_{{\lambda }_j}(T)$. Therefore [L3] identifies each projection onto $E_{{\lambda }_j}(T)$ with a polynomial in $T$.