= Solution
For a <diagonal matrix>, each diagonal entry separately obeys $d^2+2d-15=(d-3)(d+5)=0$. Hence all four possibilities are
$$
\boxed{\operatorname{diag}(3,3),\quad\operatorname{diag}(3,-5),\quad
\operatorname{diag}(-5,3),\quad\operatorname{diag}(-5,-5).}
$$
For the second form, $X^2=I$ gives
$$
(aI+bX)^2+2(aI+bX)
=(a^2+b^2+2a)I+2b(a+1)X.
$$
Because $I,X$ are linearly independent and $b\ne0$, the $X$ coefficient forces $a=-1$; the $I$ coefficient then gives $b^2=16$. Thus
$$
\boxed{A=-I\pm4X=\begin{pmatrix}-1&\pm4\\\pm4&-1\end{pmatrix}.}
$$
Both constants are nonzero, as required.
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