= Polynomials of a normal operator are normal
{title2=$f(T)^*=\bar f(T^*),\quad f(T)f(T)^*=f(T)^*f(T)$}
For a <normal operator> $T$ on a real or complex <inner product space> and a <polynomial> $f(x)=\sum_j a_jx^j$, the <adjoint operator> identity gives
$$
f(T)^*=\sum_j\bar a_j(T^*)^j.
$$
Every power of $T$ commutes with every power of $T^*$, since $TT^*=T^*T$. Expanding finite sums consequently proves that $f(T)$ commutes with $f(T)^*$, so it is a <normal operator>. The bar on $f$ means coefficientwise <complex conjugation>, and is essential for complex coefficients.
Back to article page
