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On a real or complex inner product space of finite dimension, a normal operator has a square-free polynomial as its minimal polynomial. To prove this without the spectral theorem, suppose an irreducible polynomial $f$ occurs at least twice in $m_T$. Put $h=m_T/f^2$ and $N=f(T)$. Since polynomials of a normal operator are normal, $N$ is a normal operator. For every $v$,By kernel stabilization for a normal operator, $Nh(T)v=0$. Hence the smaller-degree monic polynomial $m_T/f$ annihilates $T$, contradicting minimality.

Over the complex numbers, all irreducible polynomials are linear, so the primary decomposition theorem for linear maps now gives a basis of eigenvectors. Over the real numbers, irreducible quadratics can remain: a nontrivial planar rotation need not have any real eigenvectors, although its minimal polynomial is square-free.

For a polynomial $f(x)={\sum }_{j=0}^d{a}_j{x}^j$, the defining identity and uniqueness of the adjoint operator giveIndeed, the adjoint operator of a sum is the sum of the adjoint operators, the adjoint of $aT$ is $\bar{a}{T}^{\ast }$, and the adjoint of a product $AB$ is $B^{\ast }{A}^{\ast }$. Since the normal operator $T$ commutes with $T^{\ast }$, each power of $T$ commutes with each power of $T^{\ast }$. Multiplying the two finite sums therefore givesHence polynomials of a normal operator are normal, including constant polynomials.