Eigenvalues of a matrix and its square Ok so I messed up my last question, I'll rephrase it:
Is there a matrix $A$ of real elements, for which this holds true:
$A^2$ has more unique eigenvalues than $A$.
If not, then what about if the elements of $A$ were complex numbers?
I didn't manage to find such a matrix yet, so I tried proving that it's impossible.
I know that the eigenvalues can be calculated by constructing the characteristic polynomial for both $A$ and $A^2$ and finding the roots.
$$|A - tI| = 0$$
$$|A^2 - tI| = 0$$
But in the general case with a $n*n$ matrix the resulting polynomials are way too complicated to say anything about, so there must be another way to do it.
Thanks
 A: Are you looking for real or complex eigenvalues? If your eigenvalues have to be real then $$A=\pmatrix{0&-1\\1&0}$$ is an example. If you allow complex eigenvalues, then you may assume that $A$ is in its Jordan canoncial form. The square $A^2$ is then an upper triangle matrix where all diagonal entries are squares of eigenvalues of $A$. Hence each distinct eigenvalue of $A^2$ corresponds to at least one eigenvalue of $A$. 
A: The answer to the first question depends on what you mean by "has".
The rotation of $90^{\circ}$ about the origin in $\mathbb{R}^2$ has no eigenvalues when working over $\mathbb{R}$:
$$\left(\begin{array}{rr}
0 & -1\\
1 & 0
\end{array}\right)$$
while its square has one (repeated) eigenvalue, namely $-1$.  This phenomenon is common when working over fields that are not algebraically closed. 
However, if we consider the eigenvalues to be complex numbers (even though the original matrix has coefficients in $\mathbb{R}$), then the above is not a counterexample, since the roots of the characteristic polynomial of the rotation are $i$ and $-i$ (two distinct), while the roots of the characteristic polynomial of the square is $-1$ doubled (one root).
So let's ask:

Can the characteristic polynomial of $A^2$ have more distinct roots (in the algebraic closure of the ground field) than the characteristic polynomial of $A$?

The answer is "no." 
Over an algebraically closd field every matrix can be put in upper triangular form (e.g., via the Jordan canonical form). Thus, we may assume that $A$ is of the form
$$A=\left(\begin{array}{cccc}
\lambda_1 & * & \cdots & *\\
0 & \lambda_2 & \cdots & *\\
0 & 0 & \cdots & *\\
\vdots & \vdots &\ddots & \vdots\\
0 & 0 & \cdots & \lambda_n
\end{array}\right).$$
Therefore,
$$A^2 = \left(\begin{array}{cccc}
\lambda_1^2 & \# & \cdots & \#\\
0 & \lambda_2^2 &\cdots & \#\\
0 & 0 & \cdots & \#\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & \lambda_n^2
\end{array}\right)$$
and the set $\{\lambda_1^2,\ldots,\lambda_n^2\}$ (the roots of the characteristic polynomial of $A^2$) contains no more distinct elements than $\{\lambda_1,\ldots,\lambda_n\}$ (the roots of the characteristic polynomial of $A$), and possibly fewer.
