On this web page it is stated that for a first order linear system of two ordinary differential equations with constant coefficients, denoted
$\left\{ \begin{array}{l} {x’_1} = {a_{11}}{x_1} + {a_{12}}{x_2}\\ {x’_2} = {a_{21}}{x_1} + {a_{22}}{x_2} \end{array} \right.,$
if the real "coefficients $a_{12}$ and $a_{21}$ have the same sign, then the discriminant of the characteristic equation will always be positive and, therefore, the roots will be real and distinct." I am having trouble proving this statement.
Here is the characteristic equation for reference:
${{{\lambda ^2} – \left( {{a_{11}} + {a_{22}}} \right)\lambda }+{ \left( {{a_{11}}{a_{22}} – {a_{12}}{a_{21}}} \right) = 0.}}$
I take the quoted statement mean that if $a_{12}a_{21}>0$ then $b^2-4c >0$, where $b=-(a_{11}+a_{22})$ and $c=-(b+a_{12}a_{21})$. Writing the discriminant in terms of $a_{12}$ and $a_{21}$ but maintaining $b$ for clarity yields $b^2+4b+4a_{12}a_{21}$.
I understand that having $a_{12}a_{21}>0$ will "help the discriminant be more positive", but it hardly seems sufficient to ensure that $b^2+4b+4a_{12}a_{21}>0$, because the $4b=-(a_{11}+a_{22})$ term may be positive or negative, and for certain values of $a_{11}$ and $a_{22}$ it can outweigh the $b^2$ term. I have tried expanding $b^2+4b$ in terms of $a_{11}$ and $a_{22}$ but nothing seems to cancel and the expression still seems like it could take negative values. What am I missing here - how can one prove the quoted statement? Or, is the statement incorrect?
The discriminant is $$(-(a_{11}+a_{22}))^2-4(a_{11}a_{22}-a_{12}a_{21})$$ $$=(a_{11}-a_{22})^2+4a_{12}a_{21} \ge 0.$$. The discriminant can be 0 only if $a_{11}=a_{22}$ and ($a_{12}=0$ or $a_{21}=0$). The staement is true if 'have the same sign' is interpreted to mean 'both positive or both negative.' If 'have the same sign' is interpreted to mean 'both positive or both negative or both 0' the statement is false and the 2x 2 identity matrix $I$ is a counter-example.