$3\times 3$ matrix with no real eigenvalues I was asked this question on my hw along with any $2\times2$ matrix with no real eigenvalue and any $4\times4$ matrix with no real eigenvalue. I got the $2\times2$ which is 
$$
\begin{bmatrix}
1 & 2 \\
-1 & -1 \\ 
\end{bmatrix}
$$
can someone help me on the $3 \times 3$ and $4\times 4$, it seems really simple but I'm kind of stuck, thanks!
 A: Assuming you're talking about matrices with real entries: any nonconstant cubic polynomial with real coefficients has a real root, by the Intermediate Value Theorem.
A: One way to solve this is to use the Frobenius companion matrix.
Given some equation:
$$P(\lambda) = \lambda^4 + c_3\lambda^3 + c_2\lambda^2 + c_1\lambda + c_0$$
You can construct a matrix that has that characteristic polynomial:
$$\begin{bmatrix}
 0 & 0 & 0 & -c_0 \\
 1 & 0 & 0 & -c_1 \\
 0 & 1 & 0 & -c_2 \\
 0 & 0 & 1 & -c_3 \\
\end{bmatrix}$$
So just choose roots that are imaginary but have pair with a complex conjugate so you get a real polynomial:
$$P(\lambda) = (\lambda - a + bi)(\lambda - a - bi)(\lambda - c + di)(\lambda - c + di)$$
$$\begin{bmatrix}
 0 & 0 & 0 & -(b^2 + a^2)(d^2 + c^2) \\
 1 & 0 & 0 & 2(ac^2 + ad^2 +  a^2c + b^2c) \\
 0 & 1 & 0 & -4ac - a^2 - b^2 - c^2 - d^2 \\
 0 & 0 & 1 & 2(a + c) \\
\end{bmatrix}$$
As long as $b \ne 0$ and $d \ne 0$ you'll have a whole lot of matrices without real eigenvalues.
A: For a $4$ x $4$ matrix, why not let $$J = \begin{pmatrix}
0&-1&0&0 \\ 
1&0&0&0 \\ 
0&0&0&-1 \\ 
0&0&1&0 
\end{pmatrix}?$$
Then, what are the eigenvalues? What are the dimensions of the eigenspaces? How does the characteristic polynomial show up in the diagonal blocks?
In fact for any $n$ x $n$ matrix, where n is even, we can simply repeat these blocks along the diagonal.
A: Hint: the eigenvalues are roots to the characteristic polynomial.
