Question

If a eigenvalue of A is $\lambda$, then the corresponding eigen value of A${}^{-1}$ is

• A. $-\lambda$
• B. $\frac{1}{\lambda }$
• C. $\lambda$
• D. $\frac{\overline{\lambda }}{{\lambda }^2}$

Answer 1

The correct option is B $\frac{1}{\lambda }$

Let A be an eigen value of A and X be a corresponding eigen vector. Then,
$AX=\lambda X$
or $X=A^{-1}\left(\lambda X\right)=\lambda \left(A^{-1}X\right)$
or $\frac{1}{\lambda }X=A^{-1}X$ [$\because$ A is nonsingular $\Rightarrow \lambda \ne 0$]
or $A^{-1}X=\frac{1}{\lambda }X$
Therefore, 1/$\lambda$ is an eigen value of A${}^{-1}$ and X is the corresponding eigen vector.