= Minimal polynomial of a normal operator
{title2=$m_T=\prod_i f_i\quad(TT^*=T^*T)$}
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 $v$,
$$
N^2h(T)v=m_T(T)v=0.
$$
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.
Back to article page
