= Solution
Let $Av=\lambda v$ for a real <eigenvalue> $\lambda$ and a nonzero real <eigenvector> $v$. Part (b) gives $\lambda\ne0$ and $A^{-1}v=\lambda^{-1}v$. The expression for $B$ in part (c) therefore gives
$$
Bv=(A+I+A^{-1})v=(\lambda+1+\lambda^{-1})v.
$$
Consequently \b[$\lambda+1+\lambda^{-1}$ is a real <eigenvalue> of $B$], with the same <eigenvector> $v$. For a real <matrix> and real <eigenvalue>, a nonzero real vector can be chosen in the real <kernel> of $A-\lambda I$; no complex-vector assumption is needed.
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