With these two equations, how do I show that either a,b,c,d must be negative, if v is not 0?

Question

If I have the equations

$$ad-bc = u^2 +v^2$$

$$a+d = 2u$$

and I want $a, b, c, d \ge 0$, then how I can show that this is impossible, if $v \ne 0$?

I.e., if $v \ne 0$, then one of $a,b,c,d$ must be negative.

My attempt:

I think that I am incorrectly assuming that I can let $a=d=u$ and then in the first equation, I subtract off $u^2$ from both sides and am left with $-bc = v^2$, from which I say that if $v\ne 0$, then either $b$ or $c$ is negative. But I don't think this is the right approach.

Any help is appreciated.

Thanks,

Answer 1

Suppose that $b,c\ge 0$.

Since $u=(a+d)/2$, we have $$ad-bc=\left(\frac{a+d}{2}\right)^2+v^2,$$ Multiplying the both sides by $4$ gives $$4ad-4bc=a^2+2ad+d^2+4v^2$$ i.e. $$-4bc=(a-d)^2+4v^2$$ The LHS is non-positive since $b,c\ge 0$, and the RHS is positive since $v\not=0$. This is a contradiction.

Hence, either $b$ or $c$ has to be negative.

Answer 2

Hint: Multiply the second equation by $a$ (or $d$, it doesn't matter) and substitute into the first equation.