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Exercise 5.A.11

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Proof. Since degree of a polynomial always decreases after applying $T$ , the polynomials $p$ with $\deg p \geq 1$ can never be eigenvalues. If $\deg p = 0$ (i.e. $p$ is a constant), then $Tp = 0$ , so the only eigenvalue of $T$ is $0$ . □

celiopassos

2017-10-06 00:00

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Is it valid for 0 to be an eigenvalue cause then T-\LamdaI would be an injective map as ker(T-\LamdaI)={0}?

vedanshivaghela • 2025-09-06

Exercise 5.A.11

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