= Primary decomposition theorem for linear maps
{title2=$V=\bigoplus_i\ker p_i(T)^{a_i}$}
= Primary decomposition theorem
{synonym}
For a <linear map> $T:V\to V$ on a <finite-dimensional vector space> over a <field> $F$, factor its <minimal polynomial> as $m_T=\prod_i p_i^{a_i}$ with distinct <monic polynomials> $p_i$ that are <irreducible polynomials>. Then
$$
\boxed{V=\bigoplus_i V_i,\qquad V_i=\ker p_i(T)^{a_i},\qquad m_{T|_{V_i}}=p_i^{a_i}}.
$$
Each $V_i$ is an <invariant subspace>, since $T$ commutes with every <polynomial> in $T$. These are canonical <vector subspaces>, determined by $T$, without requiring any <eigenvalues> in $F$.
For a proof, put $m_i=p_i^{a_i}$ and $M_i=m_T/m_i$. The <coprime polynomials> $m_i,M_i$ admit a <Bézout's identity for polynomials> $b_iM_i+c_im_i=1$. Set $E_i=b_i(T)M_i(T)$. Its image is contained in $V_i$ because $m_i(T)E_i=0$. The <polynomial> $\sum_i b_iM_i-1$ is divisible by every $m_i$, hence by their product, so $\sum_iE_i=I$. On $V_i$, $E_i=I$; on $V_j$ with $j\ne i$, $E_i=0$ because $m_j$ divides $M_i$. These facts prove the <direct sum> and identify $E_i$ as its <projection operators>. Finally, if the <minimal polynomial> of any <restriction> had a smaller exponent, $m_T/p_i$ would annihilate every summand and hence all of $V$, a contradiction.
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