If $A$ has eigenvalues $\lambda_1, \lambda_2$, then $\lambda_1 \lambda_2 \in \Bbb R$ Check whether the statement below is true or give a counterexample to show it is false.  

If $A$ is a $2 \times 2$ matrix with real entries with eigenvalues $\lambda_1$ and $\lambda_2$, then $\lambda_1\cdot\lambda_2$ is always real.

I'm pretty sure I understand it, and that it's true, I'm just not sure how to prove it.
 A: There are solutions above. I offered another solution:
We know for eigenvalue $\lambda_1$ and eigenvector $x_1$,
$$ A x_1 = \lambda_1 x_1$$
So take the complex conjugate, we have
$$ \overline{Ax_1} = \bar{A}\overline{x_1} = \overline{\lambda_1}\overline{x_1} = A\overline{x_1}$$
If an eigenvalue is a comnplex number, the conjugate $\bar{\lambda}$ is also the eigenvalue. For $2\times 2$ matrix, either both are real, or one is the complex conjugate of the other so $\lambda_1 \lambda_2 = \lambda_1 \bar{\lambda_1} = |\lambda_1|^2$
A: Quick proof that the determinant is the product of the eigenvalues: Write
$$\det(A-x I) = (\lambda_1 - x)(\lambda_2 - x) \cdots (\lambda_n - x),$$
where $A$ is an $n\times n$ matrix. (IOW, factor the characteristic polynomial to get the eigenvalues.) Substituting $x=0$,
$$\det A = \lambda_1 \cdot \lambda_2 \cdot \cdots \cdot\lambda_n.$$
A: Let $\chi = ax^2 + bx + c$ be the characteristic polynomial of $A$. Then $\chi$ is with real coefficients. Using the quadratic formula,
$$ x = \frac{-b\pm \sqrt{b^2 - 4ac}}{2a}
$$
so that $\lambda_1$ is the conjugate of $\lambda_2$, and their product is real.
I don't know if you can use the following fact or not, but it makes this question easy: The product of eigenvalues of a matrix is the determinant.
A: If $A$ 2x2 matrix the characteristic equation is
$\lambda^2 - trace(A)\lambda + \det(A)$
and if A has real entries then that is a two degree polynomial with real coefficients.  And $\lambda_1\lambda_2 = \det A.$ 
