Theorem EIM says that if \(\lambda\) is an eigenvalue of the nonsingular matrix A, then \(\frac{1}{\lambda}\) is an eigenvalue of \(A^{-1}\) Write an alternate proof of this theorem using the characteristic polynomial and without making reference to an eigenvector of A for \(\lambda\).
Since \(\lambda\) is an eigenvalue of a nonsingular matrix, $0 $ (Theorem SMZE). A is invertible (Theorem NI), and so \(-\lambda A\) is invertible (Theorem MISM). Thus \(-\lambda A\) is nonsingular (Theorem NI) and det \((-\lambda A) \neq 0\) (Theorem SMZD).
[ \[\begin{aligned} pA^{-1}(\frac{1}{\lambda}) &=det(A^{-1} - \frac{1}{\lambda}I_{n} ) && \text{(Definition CP)}\\ &=1 det(A^{-1} - \frac{1}{\lambda}I_{n} ) && \text{(Property OCN)}\\ &=\frac{1}{det(-\lambda A)}det(-\lambda A)det(A^{-1} - \frac{1}{\lambda}I_{n} ) && \text{(Property MICN)}\\ &=\frac{1}{det(-\lambda A)}det((-\lambda A)(A^{-1} - \frac{1}{\lambda}I_{n})) && \text{(Theorem DRMM)}\\ &= \frac{1}{det(-\lambda A)}det(-\lambda AA^{-1}-(-\lambda A)\frac{1}{\lambda}I_{n}) && \text{(Theorem MMDAA)}\\ &= \frac{1}{det(-\lambda A)}det(-\lambda I_{n} -(-\lambda A)\frac{1}{\lambda}I_{n}) && \text{(Defination MI)} \\ &= \frac{1}{det(-\lambda A)}det(-\lambda I_{n} + \lambda A\frac{1}{\lambda}I_{n} && \text{(Theorem MMSMM)}\\ &= \frac{1}{det(-\lambda A)}det(-\lambda I_{n} + AI_{n}) && \text{(Property MICN)}\\ &= \frac{1}{det(-\lambda A)}det(-\lambda I_{n} + A) && \text{(Theorem MMIM)}\\ &= \frac{1}{det(-\lambda A)}det(A -\lambda I_{n} ) \\ &= \frac{1}{det(-\lambda A)}pA(\lambda) && \text{(Definition CP)}\\ &= \frac{1}{det(-\lambda A)}0 && \text{(Theorem EMRCP)}\\ &= 0 \end{aligned}\]]
So \(\frac{1}{\lambda}\) is a root of the characteristic polynomial of $A^{-1}??? $ and so is an eigenvalue of \(A^{-1}\)