# 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.

• What is the product of the eigenvalues of a matrix in general? – Christopher Carl Heckman Mar 28 '16 at 3:26

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 $\lambda_1 \lambda_2 = \lambda_1 \bar{\lambda_1} = |\lambda_1|^2$

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 $x=0$, $$\det A = \lambda_1 \cdot \lambda_2 \cdot \cdots \cdot\lambda_n.$$

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.

• The product of the eigenvalues of a matrix is NOT the product of the elements on its diagonal: $\pmatrix{0&1\cr 1&0\cr}$ has eigenvalues $\lambda=1,-1$. – Christopher Carl Heckman Mar 28 '16 at 3:29
• Umm... The product of the eigenvalues of a matrix is the determinant. The sum of the eigenvalues of a matrix is the sum of the elements on its diagonal (called the trace). – JMoravitz Mar 28 '16 at 3:29
• @CarlHeckman Thanks. It is corrected. – Henricus V. Mar 28 '16 at 3:30

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 $\lambda_1\lambda_2 = \det A.$