= Solution
Part (a) makes each $\lambda_j$ real. In the <eigenvector> basis,
$$
(A+iaI)e_j=(\lambda_j+ia)e_j,
$$
and $|\lambda_j+ia|^2=\lambda_j^2+a^2>0$. Hence $\boxed{A+iaI\text{ is invertible}}$.
All factors that are rational functions of $A$ commute. Taking the <Hermitian conjugate> reverses product order and gives
$$
V^\dagger=(A-iaI)^{-1}(A+iaI).
$$
Therefore
$$
V^\dagger V=(A-iaI)^{-1}(A+iaI)(A-iaI)(A+iaI)^{-1}=I,
$$
so $\boxed{V\text{ is unitary}}$. This is a <Cayley transform of a Hermitian matrix>, with scale $a$ and the convention that an infinite real eigenvalue maps to one.
Applying it to each <eigenvector> yields
$$
\boxed{Ve_j=\nu_je_j,\qquad\nu_j=\frac{\lambda_j-ia}{\lambda_j+ia}.}
$$
The <eigenvectors> are unchanged, and $|\nu_j|=1$ because numerator and denominator are <complex conjugates>. No $\nu_j$ equals one when $a>0$. The scalar map is injective on real $\lambda$, so the <eigenspaces>, including any degeneracies, are preserved exactly.
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