= Solution
Decompose $\mathbb R^n=\operatorname{span}\{a\}\oplus a^\perp$. The <rank-one matrix update> $aa^T$ acts as multiplication by $\|a\|^2$ along $a$ and as zero on its <orthogonal complement>. Thus
$$
\boxed{\lambda_\parallel(A^\pm)=1\pm\|a\|^2,\quad v_\parallel\in\operatorname{span}\{a\}\setminus\{0\};\qquad\lambda_\perp=1,\quad v_\perp\in a^\perp\setminus\{0\}.}
$$
The perpendicular eigenvalue has multiplicity $n-1$ and is absent when $n=1$. Consequently $A^+$ is always an <invertible matrix>, while
$$
\boxed{A^-\text{ is invertible if and only if }\|a\|\ne1.}
$$
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