$3\times 3$ matrix with no real eigenvalues

Question

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!

Answer 1

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.

Answer 2

Hint: the eigenvalues are roots to the characteristic polynomial.

Answer 3

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.

Answer 4

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.