Solution CB.T10 Robert Beezer

Let $\vect{v}$ be an eigenvector of $T$ for the eigenvalue $\lambda$. Then, \begin{align*} \lt{\ltinverse{T}}{\vect{v}}&= \frac{1}{\lambda}\lambda\lt{\ltinverse{T}}{\vect{v}}&&\text{$\lambda\neq 0$}\\ &=\frac{1}{\lambda}\lt{\ltinverse{T}}{\lambda\vect{v}}&&\knowl{./knowls/theorem.ILTLT.knowl}{\text{Theorem ILTLT}}\\ &=\frac{1}{\lambda}\lt{\ltinverse{T}}{\lt{T}{\vect{v}}}&&\text{$\vect{v}$ eigenvector of $T$}\\ &=\frac{1}{\lambda}\lt{I_V}{\vect{v}}&&\knowl{./knowls/definition.IVLT.knowl}{\text{Definition IVLT}}\\ &=\frac{1}{\lambda}\vect{v}&&\knowl{./knowls/definition.IDLT.knowl}{\text{Definition IDLT}} \end{align*} which says that $\displaystyle\frac{1}{\lambda}$ is an eigenvalue of $\ltinverse{T}$ with eigenvector $\vect{v}$. Note that it is possible to prove that any eigenvalue of an invertible linear transformation is never zero. So the hypothesis that $\lambda$ be nonzero is just a convenience for this problem.