 Solution

Put $p(t)={\prod }_{j=1}^m(t-{\mu }_j)$ and define the Lagrange interpolation polynomialsTheir denominators are nonzero and ${\ell }_j({\mu }_i)={\delta }_{ij}$. The polynomial ${\sum }_j{\ell }_j-1$ has degree at most $m-1$ and vanishes at all $m$ distinct scalars, so it is zero. Suppose condition (a) holds, and put $E_j={\ell }_j(T)$. Since $(t-{\mu }_j){\ell }_j(t)$ is a constant multiple of $p(t)$,Thus each $E_{j}v$ lies in $V_j$, and every vector is a sum of such vectors. If ${\sum }_j{v}_j=0$ with $v_j\in V_j$, applying $E_i$ gives $v_i=0$, because ${\ell }_i(T)v_j={\ell }_i({\mu }_j)v_j$. Hence the sum is direct, proving (b). Conversely, on each $V_j$ the factor $T-{\mu }_{j}I$ vanishes; the factors commute, so $p(T)$ vanishes on every summand and therefore on $V$. This proves the equivalence without invoking primary decomposition.

If $ST=TS$ and $v\in V_j$, then $(T-{\mu }_{j}I)Sv=S(T-{\mu }_{j}I)v=0$, so $S[V_j]\subseteq V_j$. Conversely, if each summand is preserved, then on $V_j$ both $ST$ and $TS$ act as ${\mu }_{j}S$. They agree on all summands and hence on $V$. Therefore