I am given orthogonal linear transformation $U: \mathbb{R}^4 \to \mathbb{R}^4$, represented by
$$A=\begin{bmatrix} 1/2 & 1/2 & 1/2 & -1/2 \\ 1/2 & 1/2 & -1/2 & 1/2 \\ 1/2 & -1/2 & 1/2 & 1/2 \\ 1/2 & -1/2 & -1/2 & -1/2 \end{bmatrix}$$
I want to find the characteristic and minimal polynomials of $(A+A^{-1})$ and $A$. Can I somehow use the char. and min. polynomials of the former to compute the latter? I am also required to express $A$ as a block diagonal matrix, where the blocks are $1$ and $2$ dimensional reflections and rotations, representing $U$ wrt some orthonormal basis in $\mathbb{R}^4$.
Since $A$ is indeed orthogonal, as you can easily check, $A^{-1}$ coincides with the transpose of $A$, and it is then easy to compute $$A+A^{-1}=\pmatrix{1&1&1&0\cr 1&1&-1&0\cr 1&-1&1&0\cr 0&0&0&-1}=M.$$ From the last column it is clear the minimal polynomial of $M$ has a factor $X+1$, and the image of $M+I$ is the subspace $W$ spanned by $v_1=(2,1,1,0)$, and $v_2=(1,2,-1,0)$ (its two first columns); the restriction to$~W$ of (the action of) $M$ has matrix $\binom{2~~0}{0~~2}$, since $Mv_1=2v_1$ and $Mv_2=2v_2$. So $W$ is in fact the eigenspace of $M$ for the eigenvalue $2$, and $M$ is diagonalisable with minimal polynomial $\mu=(X+1)(X-2)$, and characteristic polynomial $\chi=(X+1)^2(X-2)^2=\mu^2$, since both eigenspaces are of dimension$~2$.
For the minimal and characteristic polynomials of $A$ itself it is now easiest to consider the restrictions of its actions to the two eigenspaces of $M$ (since $A$ and $M$ clearly commute, those eigenspaces are $A$-stable). On the basis $[v_1,v_2]$ of the eigenspace of $M$ for $\lambda=2$ one obtains $\binom{1~~0}{0~~1}$ as matrix of the restriction of $A$; for the eigenspace $\ker(M+I)$ of $M$ for $\lambda=-1$ one can choose the basis $[v_3,v_4]$ where $v_3=(-1,1,1,0)$ and $v_4=(0,0,0,1)$, and the restriction of $A$ to that eigenspace on that basis is $\binom{-1/2~~~1/2}{-3/2~~~-1/2}$, whose characteristic polynomial $X^2+X+1$ being square-free (indeed irreducible in $\Bbb R[X]$) is also its minimal polynomial (the latter can also be computed by computing the relation $v_4+Mv_4+M^2v_4=0$). Then the minimal polynomial of $A$ is $(X-1)(X^2+X+1)$ and its characteristic polynomial is $(X-1)^2(X^2+X+1)$.
In passing one has found the block matrix form of $A$ after basis change to $[v_1,v_2,v_3,v_4]$, namely with the two $2\times2$ blocks mentioned above (which is in fact a $3$-block diagonal form with the two first blocks equal to $(1)$, a $1\times1$ block). For the final block other forms are possible (for appropriate bases), such as the companion matrix $\binom{0~~-1}{1~~-1}$ of $X^2+X+1$.)