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https://texercises.com/exercise/characteristic-polynomial-and-injectivity/
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Exercise:
Let V be a vector space over K and T:Vrightarrow V a linear map. Then lambda in K is an eigenvalue of T iff the linear map T-lambda id_V is not injective.

Solution:
Proof. lambda textis an eigenvalue of T &iff exists neq vin V:Tvlambda v &iff exists neq vin V:Tv-lambda v &iff exists neq vin V:T-lambda id_Vv &iff textKerT-lambda id_Vneq &iff T-lambda id_V textis not injective.
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Exercise:
Let V be a vector space over K and T:Vrightarrow V a linear map. Then lambda in K is an eigenvalue of T iff the linear map T-lambda id_V is not injective.

Solution:
Proof. lambda textis an eigenvalue of T &iff exists neq vin V:Tvlambda v &iff exists neq vin V:Tv-lambda v &iff exists neq vin V:T-lambda id_Vv &iff textKerT-lambda id_Vneq &iff T-lambda id_V textis not injective.
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characteristic polynomial, eigenvalue, eigenvector, eth, fs23, injective, lineare algebra, proof
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(3, default)
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0 (default)
Language
ENG (English)
Type
Proof
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