Let $A$ be a real $4\times4$ matrix such that $A^2+A+I = 0$. Can $A$ be orthogonal? I'm struggling on the third point of this exercise:
Let $A$ be a $4 \times 4$ matrix such that $A^2 + A + I=0$. Are this statements true or false?


*

*$A$ is invertible

*$A$ can be skew-symmetric

*$A$ can be orthogonal


This is what i have done so far:
$A^2 + A = -I \Rightarrow det(A^2+A)=det(A) \cdot det(A+I) = 1$ hence $det(A) \neq 0$ so the first statement is true.
For the second i noticed that $-1$ is an eigenvalue for $A^2+A$ and its geometric multiplicity is $4$ so also the algebraic multiplicity is $4$. Hence $-1$ is the only eigenvalue of $A^2+A$. Now let $\lambda$ be an eigenvalue of $A$ and $v \neq 0$ be a $\lambda$-eigenvector:
$$(A^2+A)v = A^2v+Av=\lambda^2v+\lambda v = (\lambda^2 +\lambda)v$$ 
hence $\lambda^2+\lambda$ is an eigenvalue of $A^2+A$ so $\lambda^2+\lambda=-1 \Rightarrow \lambda = -\displaystyle\frac{1}{2} \pm \frac{\sqrt3}{2}i $. Since skew-symmetric matrices has only pure imaginary eigenvalues the second statement is false.
For the third point i have seen that $det(A) = \left(-\displaystyle\frac{1}{2} + \frac{\sqrt3}{2}i\right)\left(-\displaystyle\frac{1}{2} - \frac{\sqrt3}{2}i\right) = 1$ and also that $\mid\lambda\mid = 1$ thus i believe that the third is true but i can't find an example of an orthogonal matrix.
Am i missing something? Have i made any mistake in my reasoning? Thank you very much in advance!
 A: Hint: a real $2 \times 2$ matrix representation of $-\frac{1}{2} + \frac{\sqrt{3}}{2} i$ would be
$$ \begin{bmatrix} -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\
\frac{\sqrt{3}}{2} & -\frac{1}{2} \end{bmatrix}. $$
Check that this matrix satisfies $A^2 + A + I = 0$ and is orthogonal.  Now, can you think of a way to use this to create a $4 \times 4$ example?
A: You need only find a $2 \times 2$ example, say $B$ and then create $A= \operatorname{diag}(B,B)$.
What are the eigenvalues of $\begin{bmatrix} -\cos t & -\sin t \\ \sin t & -\cos t \end{bmatrix}$?
A: Remark that $x^3-1=\left(x-1\right)\left(x^2+x+1\right)$
Hence you have the equality
$$
A^3=I_n
$$
Suppose that $A$ satisfies $^{t}AA=A^{t}A$ then it is orthogonal because
$^{t}AA$ is symmetric
$$
\left(^{t}AA\right)^3=I \Leftrightarrow PD^3P^{-1}=I=PP^{-1}
$$
where $D$ is diagonalizable hence
$$
D^3=I \text{ so }D=I
$$
Hence
$$
^{t}AA=I
$$
It may should help you ( nothing says that such a matrix exists ).
A: Obviously $A(-A-I)=-A^2-A=I$ therefore $A^{-1}=-A-I$ and $A$ is invertible.
For being skew symmetric we must have $A^t=-A$ then:
$$A^{-1}=-A-I\to (A^{-1})^t=(A^{t})^{-1}=-A^t-I\to -A^{-1}=-I+A$$
adding up two equations we get to
$$A^{-1}-A^{-1}=(-I-A)+(-I+A)\to 2I=0$$
which is a contradiction. Then $A$ can't be skew symmetric.
For orthogonality note that $AA^t=I$ or $A^t=A^{-1}$. then by substituting we 
obtain:
$$A^t=-A-I$$
Also if $A=[a_{ij}]_{n\times n}$ then $A^t=[a_{ji}]_{n\times n}$ and if two matrices are equal so are their pairwise entries which leads us to:
if $i\ne j$ then $a_{ji}=-a_{ij}$ and if $i=j$ then $a_{ii}=-1-a_{ii}\to a_{ii}={-1\over 2}$ so for orthogonality the matrix must be skew symmetric which is a contradiction. So the matrix can neither be skew symmetric nor orthogonal.
