Polynomials of a normal operator are normal ($f(T{)}^{\ast }=\bar{f}(T^{\ast }),f(T)f(T{)}^{\ast }=f(T{)}^{\ast }f(T)$)

For a normal operator $T$ on a real or complex inner product space and a polynomial $f(x)={\sum }_j{a}_j{x}^j$, the adjoint operator identity givesEvery power of $T$ commutes with every power of $T^{\ast }$, since $T{T}^{\ast }=T^{\ast }T$. Expanding finite sums consequently proves that $f(T)$ commutes with $f(T{)}^{\ast }$, so it is a normal operator. The bar on $f$ means coefficientwise complex conjugation, and is essential for complex coefficients.