Question #213794

Suppose T \isin L(V) is invertible.

(a) Suppose λ\lambda\isin F with λ\lambda not= 0. Prove that λ\lambda is an eigenvalue of T if and only if 1/λ\lambda is an eigenvalue of T-1.

(b) Prove that T and T-1 have the same eigenvectors.


1
Expert's answer
2021-07-07T11:14:25-0400

(a)

λ\lambda is an eigenvalue of TT

    \iff there exists a nonzero vVv\in V such that Tv=λvTv=\lambda v

    \iff there exists a nonzero vVv\in V such that T1Tv=T1λvT^{-1}Tv=T^{-1}\lambda v

    \iff there exists a nonzero vVv\in V such that v=λT1vv=\lambda T^{-1}v

    \iff there exists a nonzero vVv\in V such that 1λv=T1v\dfrac{1}{\lambda}v=T^{-1}v

    1λ\iff \dfrac{1}{\lambda} is an eigenvalue of T1.T^{-1}.


(b)

Note that since TT is invertible, TT is injective.

Suppose (0,v)(0, v) is an eigencouple of T.T. Then Tv=0Tv=0 ; thus v=0v=0 and in fact there cannot be any eigenvectors of TT corresponding to 0.0.

Now, suppose vv is an eigenvector of TT corresponding to λ0.\lambda\not=0.

By the proof of (a), vv is an eigenvector of T1T^{-1} corresponding to 1λ.\dfrac{1}{\lambda}.

So all eigenvectors of TT are eigenvectors of T1T^{-1}; reversing the roles of TT and T1T^{-1}yields the reverse inclusion and completes the proof.


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