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Out of Trefethen and Bau's "Numerical Linear Algebra":

For any $\mu \in \mathbb{R}$ that is not an eigenvalue of $A$, the eigenvectors of $(A-\mu I)^{-1}$ are the same as the eigenvectors of $A$, and the corresponding eigenvalues are $\{(\lambda_j-\mu)^{-1}\}$, where $\{\lambda_j\}$ are the eigenvalues of A. This suggests an idea.

And then they go and apply power iteration to $(A-\mu I )^{-1}$ and call it inverse iteration.

Fine ... except this statement's validity isn't quite as obvious to me as it apparently should be.

Why does the quoted statment hold?

(Not sure if any of these are important, here, but you should know that as a simplification, $A$ has been declared to be real and symmetric and because "most of these ideas pertain to [the iterative process for the final convergence of an eigenvalue algorithm]", "$A$ will be not just real and symmetric, but tridiagonal" "by the time we come to applying these ideas", which "is occasionally of mathematical importance".)

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Note, that if $\lambda$ is an eigenvalue of $A$, then $\lambda-\mu$ is an eigenvalue of $A-\mu I$. This can easily seen, evaluating $(A-\mu I)\cdot v$ for $Av=\lambda v$. Therefore we know the eigenvalues of $(A-\mu I)$.

Next keep in mind, that if $\lambda$ is an eigenvalue of $A$, then $\lambda^{-1}$ is an eigenvalue of $A^{-1}$. This again can be seen by evaluating $A^{-1}v$ with $v$ as eigenvector.

Combining these here, we can see, that $(\lambda - \mu)^{-1}$ is indeed an eigenvalue of $(A-\mu I)^{-1}$.

Keep in mind, that I never used symmetry, sparsity or any other information about $A$ than it being an square matrix with $Av=\lambda v$.

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    You should be a bit careful: $A^{-1}$ need not exist, and $\lambda^{-1}$ might not make sense... However under the assumption that $\mu$ is not an eigen-value of $A$, $A-\mu I$ is (by definition) invertible... – 2017-01-18
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    And the same equations show the identical eigenvectors. Thank you! – 2017-01-18