Triangularization requires the characteristic polynomial to split

Created almost 9 years ago, updated about 8 hours ago

An operator on a finite-dimensional space is triangularizable over $\mathbb F$ iff its characteristic polynomial splits over $\mathbb F$. In particular, every complex operator is triangularizable, but not every real one is: a planar $90^\circ$ rotation has no real eigenvalue. Triangularizability is weaker than diagonalizability.

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