real matrix with non-simple pure imaginary eigenvalue I am looking for a matrix of non-simple pure imaginary eigenvalue. Obviously, it should be an $n\times n$ matrix where $n >2$. However, I could not find such a matrix. Can anyone tell me one or how to construct one?
 A: Hint: If you can find a matrix $A$ that has pure imaginary eigenvalues, then the matrix $\begin{pmatrix}A & 0 \\ 0 & A\end{pmatrix}$ will have those same eigenvalues repeated twice (hence, not simple).
A: We use the fact that the determinant of block matrix is very easy to compute by means of the determinant of its blocks, that means:
$$
\det \begin{pmatrix}A & 0 \\ 0 & B\end{pmatrix}=\det(A)\det(B)
$$
for suitable $A,B$. You can see a nice prove of this fact here.
Now we just need to find a proper $A$ and set our matrix to be 
$$
S=\begin{pmatrix}A & 0 \\ 0 & A\end{pmatrix}
$$
The easiest way to find such a matrix $A$ is to reconsruct the characteristic polynomial $\chi_A(\lambda)$:
So we want to find something like 
$$
\chi_A(\lambda)=\lambda^2+1=0\Leftrightarrow \lambda_{1,2}=\pm i
$$
this polynomial is represented by the following matrix $A$
$$
A=\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}
$$
That means we have constructed a matrix $S$ with pure imaginary eigenvalues of algebraic multiplicity of $2$, $\lambda_{1,2}=+i;\lambda_{3,4}=-i$
A: Any matrix of real numbers will have a characteristic polynomial of real coefficients. Any polynomial of real coefficients will have zeros which are either real or come in complex conjugated pairs. So you won't be able to escape having a $-i$ eigenvalue to partner with $i$.
However if you consider the matrix 
$$ 
{G_{C_4}} = \left( \begin{array}{cccc}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 \\
\end{array} \right) 
$$
it will be such as $(G_{C_4})^4 = I$ which is the property of the imaginary unit if we consider the matrix $G_{C_4}$ to "represent" $i$ and the $I$ matrix to "represent" $1$.
