Solutions of $A^2+A^t=I_n$ in $M_n(\mathbb{R})$ I would like to prove that, if $A\in M_n(\mathbb{R})$ satisfies
$$A^2+A^t=I_n$$
then neither 0 nor 1 can be in $Sp(A)$. 
Such an $A$ satisfies $A(A-I)(A^2+A-I_n)=0$, hence $Sp(A)\subset \{0;1;a;b\}$, where $a$ and $b$ are the roots of $X^2+X-1$, therefore $A$ can be diagonalised. Moreover, as $AA^t=A^tA$, one can find a common basis of diagonalisation for $A$ and $A^t$. However, I am stuck at that point, that is I can't prove that 0 and 1 are not in $Sp(A)$, even though I can prove that:
•$x$ is a an eigenvector for $A$ for 0 iff $x$ is an eigenvector of $A^t$ for 1
• $x$ is an eigenvector for $A$ for $a$ (resp. $b$) iff $x$ is an eigenvector of $A^t$ for $a$ (resp. $b$)
which give the matrix of $A^t$ in the new basis if one has the matrix of $A$.
Incidentally, once it is proved that 0 and 1 are not eigenvalues, one easily deduces that $A$ is in fact symmetric…
 A: Let $A=QTQ^t$ be the real Schur decomposition, where $Q$ is an orthogonal matrix and $T$ is a real block upper triangular matrix, with $1\times1$ or $2\times2$ blocks on the diagonal.
Your condition is equivalent to
$$T^2+T^t=I\,.\tag{1}$$
Since $T^2$ is block upper triangular and $I-T^t$ is block lower triangular, it follows that $T$ is block diagonal. Writing $T=\operatorname{diag}(B_1,\ldots,B_k)$, and inserting it into $(1)$, we obtain that
$$B_i^2+B_i^t=I\,,\quad i=1,\ldots,k\,.$$
That is, each diagonal block also satisfies your condition.
Let $\lambda$ be any real eigenvalue of $A$. Then it is represented by $1\times 1$ block in the Schur decomposition, and consequently satisfies
$$\lambda^2+\lambda=1\,.$$
Therefore, $0$ and $1$ can not be in the spectrum of $A$.
A: Note that for any $x\in\mathbb{R}^n$ we have
$$x^tAx = (x^tAx)^t = x^tA^tx\,.\tag{1}$$
Now, assume $1$ is an eigenvalue of $A$, and let $x\neq 0$ be an associated eigenvector. 
Then $x$ is an eigenvector of $A^t$ for $0$. Our assumption leads to
$$0\neq x^tx = x^tAx = x^tA^tx = 0\,,$$
which contradicts $(1)$. Therefore, $1$ is not an eigenvalue of $A$.
The proof that $0$ is not an eigenvalue of $A$ goes analogously.
