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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.

Let $w=Tv$. If $T^{2}v=0$, then $Tw=0$, and part (i) gives $\Vert T^{\ast }w\Vert =\Vert Tw\Vert =0$. Thus $T^{\ast }w=0$ andThe positive definiteness of the inner product gives $w=0$. Since the reverse kernel inclusion always holds, we obtainThis is kernel stabilization for a normal operator.