Finite-dimensional vector space/Linear mapping/Two generalized eigenspaces/Fact

For a linear mapping ${}\varphi \colon V\rightarrow V$ on a finite-dimensional ${}K$-vector space ${}V$ and two eigenvalues ${}\lambda \neq \delta$,

the corresponding

generalized eigenspace have trivial intersection, that is

${}\operatorname {GeEig} _{\lambda }(\varphi )\cap \operatorname {GeEig} _{\delta }(\varphi )=0\,.$

Proof, Write another proof