IMC 2011 Day 1 Problem 2 - Linear Algebra I have been reading the solutions of a past IMC paper (from 2011, Day 1)
and I did not understand the solution to Problem 2 completely.
Problem 2: Does there exist a real $3$x$3$ matrix $A$, such that $tr(A)=0$ and $A^2+A^t=I$?
Official solution:
"The Anwser is NO. Suppose that $tr(A)=0$ and $A^2+A^t=I$. Taking the transpose, we have $$A=I-(A^2)^t = I - (A^t)^2 = I - (I-A^2)^2= 2A^2 - A^4,$$
$$A^4-2A^2+A=0.$$
The roots of the polynomial $x^4-2x^2+x=x(x-1)(x^2+x-1)$ are $0,1,(\frac{-1+\sqrt 5}{2}),(\frac{-1-\sqrt 5}{2})$ so these numbers can be the eigenvalues of $A$; the eigenvalues of $A^2$ can be $0,1,(\frac{1+\sqrt 5}{2}),(\frac{1-\sqrt 5}{2})$.
By $tr(A)=$, the sum of the eigenvalues of $A$ is $0$, and by $tr(A^2)=tr(I-A^t)=3 the sum of squares of the eigenvalues is 3. It is easy to check, that this two conditions cannot be satisfied simultaneously."
It says, that if the only possible eigenvalues for a real $3$x$3$ matrix $A$ are $0,1,(\frac{-1+\sqrt 5}{2}),(\frac{-1-\sqrt 5}{2})$, then the only possible eigenvalues for $A^2$ are $0,1,(\frac{1+\sqrt 5}{2}),(\frac{1-\sqrt 5}{2})$. 
But why is this so? Shouldn´t it be, that if $\lambda_{i}$ are the eigenvalues for $A$, then $\lambda_{i}^2$ are the eigenvalues for $A^2$? But if the latter is true and they made an mistake, the only possible eigenvalues for $A^2$ are $0,1,(\frac{3+\sqrt 5}{2}),(\frac{3-\sqrt 5}{2})$ and their argument fails, since you can take the eigenvalues to be $0,(\frac{3+\sqrt 5}{2}),(\frac{3-\sqrt 5}{2})$ so that $tr(A^2)=3$ and there is no contradiction.
Or am I missing something (is there a theorem, etc. that supports their reasoning)?
Please help
 A: So first of all, you're right, they made a mistake, in the stated setting, we have for the eigenvalues of $A$ the possible set of 
$$\left\{0,1,(\frac{-1+\sqrt 5}{2}),(\frac{-1-\sqrt 5}{2})\right \}
$$
which gives us, since for $\lambda$ being an eigenvalue to the eigenvector $v$ of $A$
$$
A(Av)=A\lambda v=\lambda Av=\lambda^2v
$$
the following set of possible eigenvalues for $A^2$
$$\left\{0,1,(\frac{3+\sqrt 5}{2}),(\frac{3-\sqrt 5}{2})\right \}
$$
Now we have the restrictions $\operatorname{tr}(A)=0 \tag 1$ and $\operatorname{tr}(A^2)=3 \tag 2$
which arises from $A^2+A^t=I$. Assume now we only have $0$ eigenvalues, then $(1)$ holds but  $(2)$ doesn't. So we can exclude this.
Assume now we take $(\frac{-1+\sqrt 5}{2})$, then we also have to take $(\frac{-1-\sqrt 5}{2}),1$. Because otherwise we never fulfill $(1)$. If we take any of $\left\{1,(\frac{-1+\sqrt 5}{2}),(\frac{-1-\sqrt 5}{2})\right \}$, we always have to take all three of them.
This means, that the eigenvalues of $A^2$ look in this case like
$$\left\{1,(\frac{3+\sqrt 5}{2}),(\frac{3-\sqrt 5}{2})\right \}$$
but because of $(2)$ we need to satisfy
$$
\operatorname{tr}(A^2)=3 \text{ but } \operatorname{tr}(A^2)=1+\frac{3+\sqrt 5}{2}+\frac{3-\sqrt 5}{2}\equiv4
$$
and therefore $(2)$ can never hold, which makes it impossible to have a real $3\times3$ matrix $A$, such that $\operatorname{tr}(A)=0$ and $A^2+A^t=I$.
