Prove that the eigenvectors of $(A - pI)^{-1}$ are the same as the eigenvectors of $A$ for real, symmetric $A$ From the book "Numerical Linear Algebra" p. 206:

$A$ is a real, symmetric matrix. For any $p\in R$ that is not an eigenvalue of $A$, the eigenvectors of
  $(A - pI)^{-1}$ are the same as the eigenvectors of $A$, and the
  corresponding eigenvalues  are$ \{(q_j - p)^{-1}\}$, where $\{q_j\}$
  are the eigenvalues of $A$

How is this result derived?
 A: $$Av=\lambda v \\
\iff (A-pI)v=(\lambda-p)v \\
\iff (A-pI)^{-1}v=(\lambda-p)^{-1}v
$$
Since $p$ is not an eigenvalue of $A$, $A-pI$ is invertible.  We assume $(\lambda,v)$ an eigenpair of $A$ on the top, and $v$ to be an eigenvector at the bottom.
A: Actually this is a combination of a few different results that are more easily proven separately but just for this I will prove everything in one go.
Suppose $e_i$ is an eigenvector of $(A - pI)^{-1}$ then
\begin{align}
(A - pI)^{-1}e_i&=(q_i - p)^{-1}e_i\\
(A - pI)(A - pI)^{-1}e_i&=(A - pI)(q_i - p)^{-1}e_i\\
(q_i - p)e_i&=(A - pI)e_i\\
q_ie_i - pe_i&=Ae_i - pe_i\\
 Ae_i&=q_ie_i\\
\end{align}
Clearly, $e_i$ is an eigenvector of $A$ with corresponding eigenvalue $q_i$.

The results I used in this derivation:

If $\lambda$ is an eigenvalue of $M$ with corresponding eigenvector $v$, then $\lambda^{-1}$ is an eigenvalue of $M^{-1}$ with corresponding eigenvector $v$, provided $M$ is invertible.

\begin{align}
Mv&=\lambda v\\
M^{-1}Mv&=M^{-1}\lambda v\\
Iv&=\lambda M^{-1}v\\
\lambda^{-1}v&=M^{-1}v
\end{align}

If $\lambda$ is an eigenvalue of $M$ with corresponding eigenvector $v$, then $\lambda+k$ is an eigenvalue of $M+kI$ with corresponding eigenvector $v$.

\begin{align}
Mv&=\lambda v\\
(M+kI)v&=Mv+kIv\\
&=\lambda v+kv\\
&=(\lambda+k)v
\end{align}
Bonus:

If $\lambda$ is an eigenvalue of $M$ with corresponding eigenvector $v$, then $n\lambda$ is an eigenvalue of $nM$ with corresponding eigenvector $v$.

\begin{align}
Mv&=\lambda v\\
nMv&=n\lambda v\\
\end{align}
A: If $(A-pI)=-p(I-A/p)$ is invertible then $[-p(I-A/p)]^{-1}=-\frac{1}{p}\left[I-\frac{A}{p}+\frac{A^2}{p^2}+\frac{A^3}{p^3}+\cdots\right]$. If $v$ is an eigenvector of $A$ corresponding to the eigenvalue $q$, then $(A-pI)^{-1}v=-\frac{1}{p}\left[1-\frac{q}{p}+\frac{q^2}{p^2}+\frac{q^3}{p^3}+\cdots\right]v=(q-p)^{-1}v$
A: $$(A-pI)v_i=q_iv_i-pv_i\implies (A-pI)^{-1}(A-pI)v_i=(A-pI)^{-1}(q_i-p)v_i\implies (A-pI)^{-1}v_i=(q_i-p)^{-1}v_i \quad \square$$
