 Eigenspace decomposition from a square-free annihilating polynomial

For distinct scalars ${\mu }_j$, the Lagrange interpolation polynomials ${\ell }_j(t)={\prod }_{k\ne j}(t-{\mu }_k)/({\mu }_j-{\mu }_k)$ sum to one. Their values at $T$ have images in the respective eigenspaces, since $(t-{\mu }_j){\ell }_j$ is a multiple of the annihilating polynomial. Hence every vector is the sum of its ${\ell }_j(T)$ images. On the $i$th eigenspace, ${\ell }_j(T)$ is multiplication by ${\delta }_{ij}$, which proves directness. This gives the decomposition without invoking the primary decomposition theorem. Under this decomposition a linear operator commutes with $T$ exactly when it preserves every eigenspace.