Show 1/lambda = eigenvalue of A^-1

**sparky** · February 9th 2011, 01:35 PM

I am not sure where to begin with this question but any guidance will be greatly appreciated:

Question: Let A be a nonsingular matrix and let lambda be an eigenvalue of A. Show that 1/lambda is an eigenvalue of A^-1

**FernandoRevilla** · February 9th 2011, 02:25 PM

If $A\in \mathbb{K}^{n\times n}$ is nonsingular and $\lambda \in \mathbb{K}$ is an eigenvalue of $A$ then, $\lambda\neq 0$ (why?) and there exists $0\neq x \in \mathbb{K}^n$ column vector such that $Ax=\lambda x$ . Now, multiply both sides by $A^{-1}$ .

Fernando Revilla

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