Show that any conjugate pair of complex numbers (with non-zero imaginary part) cannot be the spectrum of any 2x2 matrix with real, nonnegative entries

Question

My professor showed me this in her office today but I didn't like her method and wanted to use another method.

So, I computed the characteristic polynomial of some arbitrary $2 \times 2$ matrix with nonnegative entries $a,b,c,d$, and solved for its two eigenvalues using the quadratic formula.

Then I equated the conjugate pair of complex numbers (with non-zero imaginary part) to the two eigenvalues, and am trying to derive some contradiction at this point. But, I'm stuck.

My professor had used some complexification process and showed that the realizing matrix must have certain trigonometric entries (she mentioned rotation matrices), and then derived an impossibility.

Thanks in advance,

Answer 1

I'll assume your matrix is

$$ M=\left( \begin{array}{ccc} a & b\\ c & d\\\end{array} \right). $$

The the characteristic polynomial is

$$ |M-\lambda I|=(a-x)(d-x)-bc $$

and setting the characteristic polynomial to zero yields the solution

$$ \lambda = \frac 12 \left(a + d \pm\sqrt{4bc + (a-d)^2}\right). $$

The eigenvalues are only imaginary if

$$ 4bc + (a-d)^2<0 $$

but by assumption $b>0$ and $c>0$. The second term is positive as long as $a$ and $d$ are real (or their complex parts cancel). Thus, the eigenvalues must be real.

Answer 2

Hint: Calculate the discriminant of the characteristic polynomial of $$\begin{pmatrix}a&b\\c&d\end{pmatrix}$$ after some easy calculations to be $$\frac{1}{4}\left((a-d)^2+4bc\right).$$

Remark: Complex-conjugated eigenvalues may only exist if $bc<0$ regardless of the signs of $a$ and $b$.

Answer 3

For a $2 \times 2$ matrix $A$, the characteristic equation is $x^2 - \text{Tr}(A)x + |A| = 0$. If this equation has complex roots, then the discriminant is negative. But if you assume A = [a,b;c,d], then $\text{Tr}(A)^2 - 4|A| = (a-d)^2 + 4bc \geq 0$!!