91Ó°ÊÓ

Eigenvalue:An eigenvalue of A is a scalar $\mathit{λ}$such that the equation $\mathbf{Av}\mathbf{=}\mathbf{λ\pm ¹}$ has a nontrivial solution.

Consider a$n×n$invertible matrix A and$\stackrel{â‡\P Ä}{v}$is an eigenvector of the matrix A with respect to the eigenvalue$\mathrm{λ}$ .

The objective is to find whether the vector$\stackrel{â‡\P Ä}{v}$ is eigenvector for is $A^{-1}$or not. If yes find the eigenvalue.

Since$\stackrel{â‡\P Ä}{v}$is an eigenvector of the matrix A with respect to the eigenvalue$\mathrm{λ}$ .

$A\stackrel{â‡\P Ä}{v}=\mathrm{λ}\stackrel{â‡\P Ä}{v}$

Find whether the vector$\stackrel{â‡\P Ä}{v}$is eigenvector for is$A^{-1}$or not as,

Consider

role="math" localid="1659528893977" $$\begin{gathered} A\stackrel{â‡\P Ä}{v}=\mathrm{λ}\stackrel{â‡\P Ä}{v}{A}^{-1}\left(A\stackrel{â‡\P Ä}{v}\right) \\ =A^{-1}\left(\mathrm{λ}\stackrel{â‡\P Ä}{v}\right) \\ {A}^{-1}\left(A^{-1}A\right)\stackrel{â‡\P Ä}{v}=\mathrm{λ}\left(A^{-1}\stackrel{â‡\P Ä}{v}\right)I\stackrel{â‡\P Ä}{v} \\ =\mathrm{λ}\left(A^{-1}\stackrel{â‡\P Ä}{v}\right)\sin ce{A}^{-1}A=I\mathrm{λ}\left(A^{-1}\stackrel{â‡\P Ä}{v}\right) \\ =\stackrel{â‡\P Ä}{v}{A}^{-1}\stackrel{â‡\P Ä}{v}=\frac{1}{\mathrm{λ}}\stackrel{â‡\P Ä}{v} \end{gathered}$$

Therefore, from equation (1) the vector $\stackrel{â‡\P Ä}{v}$is eigenvector for is $A^{-1}$with respect to the eigenvalue${\mathrm{λ}}^{-1}=\frac{1}{\mathrm{λ}}$.