Suppose that $V$ is finite-dimensional over an algebraically closed field, $F$. Thus, for any $T \in \mathcal L(V)$, there exists a polynomial $P \in F[t]$ such that $P(T) = 0$. Since $F$ is algebraically closed, we can write $P(t) = \prod_{i=1}^m (T - a_i)^{c_i}$, and apply primary decomposition to get: $$V = \oplus_{i=1}^m \, N(T - a_i I)^{c_i}$$ We will create another decomposition. Suppose that $\{ \, \lambda_1, \lambda_2 \ldots \lambda_r \, \}$ is the full set of eigenvalues of $T$. It can be shown separately that:
- $V$ is spanned by some set of generalized eigenvectors of $T$.
- Generalized eigenvectors associated with distinct eigenvalues form a linearly independent set.
- For each $i \in \{ \, 1, 2 \ldots r \, \}$, the set of general eigenvectors associated with $\lambda_i$ is a subspace, and equals $N(T - \lambda_i I)^n$, where $n = \dim V$.
When we take all three facts together, we have: $$V = \oplus_{i=1}^r \, N(T - \lambda_i I)^n$$ So, I am trying to reconcile the two decompositions. If the dimension of $N(T - a_i I)^{c_i}$ is greater than zero, then there is a nonzero $x$ such that $(T - a_i I)^{c_i}x = 0$. It follows that $a_i$ is an eigenvalue of $T$ and it is inside $\{ \, \lambda_1, \lambda_2 \ldots \lambda_r \, \}$. What contradiction will there be if $\dim N(T - a_i I)^{c_i} = 0$?
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$\begingroup$There is no contradiction expected if $\dim N(T - a_iI)^{c_i} = 0$ because there was no assumption that $P$ is a polynomial of minimal degree such that $P(T) = 0$.
If one does assume that $P$ is a polynomial of minimal degree such that $P(T) = 0$, then $\dim N(T - a_iI)^{c_i} = 0$ does lead to a contradiction because dividing $P$ by $(z - a_i)^{c_i}$ would give a polynomial of smaller degree that annihilates $T$.
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