= Solution
Let $J=\begin{pmatrix}0&-2\\1&0\end{pmatrix}$, so the displayed <rational canonical form> of $T$ is $J\oplus J$. Since $J^2=-2I_2$, we have $T^2=-2I_4$.
If $a=0$, then $U=bI_4$ is a <scalar matrix>. Its <invariant factors of a linear operator> are four copies of $t-b$, so its rational canonical form is
$$
\boxed{C(t-b)^{\oplus4}=bI_4\qquad(a=0).}
$$
Here $m_U=t-b$ and $c_U=(t-b)^4$, including when $b=0$.
Now let $a\ne0$ and put
$$
q(t)=(t-b)^2+2a^2=t^2-2bt+(b^2+2a^2).
$$
On each two-dimensional invariant block, $A=aJ+bI_2$ satisfies $q(A)=0$. The vectors $e_1,Ae_1$ form a <cyclic vector> basis, since their coordinate matrix is $P=\begin{pmatrix}1&b\\0&a\end{pmatrix}$ with <determinant> $a\ne0$. Directly,
$$
P^{-1}AP=
\begin{pmatrix}0&-(b^2+2a^2)\\1&2b\end{pmatrix}=C(q).
$$
Applying this basis change to both blocks gives
$$
\boxed{\operatorname{RCF}(U)=
\begin{pmatrix}
0&-(b^2+2a^2)&0&0\\
1&2b&0&0\\
0&0&0&-(b^2+2a^2)\\
0&0&1&2b
\end{pmatrix}\qquad(a\ne0).}
$$
Thus $m_U=q$ and $c_U=q^2$. In particular there are two invariant factors $q,q$, rather than a single block for $q^2$. Also $q(t)>0$ for every real $t$, so it is an <irreducible polynomial> over $\mathbb R$. This is an instance of <rational canonical form under an affine change>.
Finally $T^2=-2I_4$ implies $T^{-1}=-T/2$. Substituting $a=-1/2$, $b=0$ in the preceding result gives $q(t)=t^2+1/2$, and hence
$$
\boxed{\operatorname{RCF}(T^{-1})=
\begin{pmatrix}
0&-1/2&0&0\\
1&0&0&0\\
0&0&0&-1/2\\
0&0&1&0
\end{pmatrix}.}
$$
Its minimal polynomial is $t^2+1/2$ and its characteristic polynomial is $(t^2+1/2)^2$. The normalized reciprocal of $t^2+2$ is also $t^2+1/2$, agreeing with <rational canonical form of an inverse>.
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