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If $\lambda_1,\lambda_2...\lambda_n$ are eigen values of A then show that $\frac{1}{\Lambda_1} ,\frac{1}{\lambda_2}...$Are eigen values of $A^{-1}$

written 6.7 years ago by teamques10 ★ 70k

•   modified 5.7 years ago

engineering mathematics

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written 6.7 years ago by teamques10 ★ 70k

•   modified 6.7 years ago

If $\lambda$ is an eigen value of A and X is the corresponding eigen vector then

$ \begin{aligned} AX &= \lambda X \\ X&= (A)^{-1}(\lambda X)\\ X&= \lambda(A^{-1}X) \\ \frac{1}{\lambda} X &= A^{-1} X \\ \end{aligned}$

By comparing both sides $\frac {1}{\lambda}$ is an eigen value of $A^{-1}$

Hence proved.

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