If $A$ is invertible, the eigenvalues of $A^{-1}$ are
Athe same as those of $A$
Bthe negatives of those of $A$
Creciprocals: $1/\lambda_i$
Dthe squares of those of $A$
Answer & Solution
Correct answer: C. reciprocals: $1/\lambda_i$
1. If $Av = \lambda v$ with $v \neq 0$, multiply by $A^{-1}$: $v = \lambda A^{-1}v$, so $A^{-1}v = (1/\lambda) v$.
2. (Note: $\lambda \neq 0$ since $A$ is invertible.)
3. So the eigenvalues of $A^{-1}$ are $1/\lambda_i$ where $\lambda_i$ are eigenvalues of $A$.
4. The eigenvectors are the SAME.
5. Example: $A$ with eigenvalue 4 has $A^{-1}$ eigenvalue $1/4$.
6. Other options misapply the relationship.
_Source: Sergei Treil, "Linear Algebra Done Wrong", §4.5 (Eigenvalues of $A^{-1}$, derived)._
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