 Solution

Part (a) makes each ${\lambda }_j$ real. In the eigenvector basis,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 givesThereforeso $\boxed{V\text{ is unitary}}$. 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 yieldsThe 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.