Answer to Question #213794 in Linear Algebra for Simphiwe Dlamini

Question

Suppose T $\isin$ L(V) is invertible.

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

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

Answer 1

(a)

$\lambda$ is an eigenvalue of $T$

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

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

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

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

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

(b)

Note that since $T$ is invertible, $T$ is injective.

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

Now, suppose $v$ is an eigenvector of $T$ corresponding to $\lambda\not=0.$

By the proof of (a), $v$ is an eigenvector of $T^{-1}$ corresponding to $\dfrac{1}{\lambda}.$

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