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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.

All calculations here take place in the finite field $F={\mathbb{F}}_3$. For the given matrix $S$, its characteristic polynomial isThe values $q(0),q(1),q(2)$ are $2,2,1$, so this quadratic polynomial has no polynomial root in $F$ and is an irreducible polynomial. Direct calculation givesNeither $S$ nor $q(S)$ is zero. The Cayley-Hamilton theorem says that the minimal polynomial divides $xq(x)$; its proper monic polynomial divisors cannot annihilate $S$, so $m_S=xq$.

The two kernels give the required primary decomposition theorem for linear maps explicitly:The determinant of the matrix with these three basis vectors as columns is $1$ in $F$, proving $W=W_1\oplus W_2$. To check these are invariant subspaces, writing $u=(1,2,0{)}^T$ and $z=(0,0,1{)}^T$ gives $Su=u+2z$ and $Sz=2u$. Hence the matrix of the restriction to $W_2$ in the basis $(u,z)$ iswith minimal polynomial $q$; the nonzero vector space $W_1$ has zero restriction and minimal polynomial $x$.

The primary decomposition theorem for linear maps starts with a linear map $T:V\to V$ on a finite-dimensional vector space over a field $F$. Factor its minimal polynomial aswhere the $p_j$ are distinct monic polynomials that are irreducible polynomials and $a_j\ge 1$. Define $V_j=\ker p_j(T{)}^{a_j}$. ThenThus the direct sum separates the irreducible polynomial factors into uniquely determined invariant subspaces. The statement works even when the minimal polynomial does not split into linear polynomials over $F$.

Over the complex numbers, the fundamental theorem of algebra makes every irreducible polynomial factor $f_j(x)$ equal to $x-{\lambda }_j$. Part (b)(iv) giveswith distinct eigenvalues. The primary decomposition theorem for linear maps consequently expresses $V$ as the direct sum of the actual eigenspaces $E_{{\lambda }_j}=\ker (T-{\lambda }_{j}I)$, with no higher generalized eigenspaces needed. Part (ii) makes these summands mutually orthogonal.

Apply the Gram-Schmidt process to a basis within each eigenspace. Its linear combinations stay within that eigenspace, so all resulting vectors are still eigenvectors. Their union is orthonormal by part (ii) and spans $V$ by the primary decomposition theorem for linear maps. Thusfor an orthonormal basis $\mathcal{B}$ of eigenvectors. This proves the finite-dimensional complex spectral theorem for normal operators by the requested decomposition argument.