For any linear transformation $T:\mathbb{Q}^{4} \to \mathbb{Q}^{4} $, does there always exist a non-zero T- invariant subspace?
As we know
For any linear transformation $T:\mathbb{R}^{4} \to \mathbb{R}^{4}$, there exists a non-zero T- invariant subspace.
Does completeness property cause some differences for T- invariant subspaces?
$\endgroup$1 Answer
$\begingroup$Take any $4 \times 4$ matrix of rational numbers whose characteristic polynomial is irreducible over $\Bbb{Q}$, and form a linear operator $T$ on $\Bbb{Q}$ based on this matrix. If $V \le \Bbb{Q}^4$ is some proper non-trivial invariant subspace of $T$, then $T|_V$ is an operator on $V$, whose characteristic polynomial has degree $0 < \dim(V) < 4$, and which divides the characteristic polynomial of $T$. But, this can't happen, as the characteristic polynomial of $T$ is irreducible.
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