Exercise 1.5.8

Proof. Assume that $f=(a,b,c)$ is semidefinite and not definite. Reasoning by contradiction, assume that $\mathrm{\Delta }\ne 0$.

If $f$ is positive semidefinite (and not definite), by definition 1.2.9,

Then $a=f(1,0)\ge 0$, and $c=f(0,1)\ge 0$.

We prove first that $a>0$. If $a=0$, then $f(x,y)=y(\mathit{bx}+\mathit{cy})$. Then $b\ne 0$, otherwise $\mathrm{\Delta }=0$.

If $b>0$, $f(-\frac{c}{b}-1,1)=-b<0$, and if $b<0$, $f(-\frac{c}{b}+1,1)=b<0$. This is a contradiction, since $f$ is positive semidefinite. Therefore $a>0$.

Then, knowing that $f$ is positive semidefinite,

Therefore $f(-b,2a)=-4a^2\mathrm{\Delta }\ge 0$, thus $\mathrm{\Delta }\le 0$. Under the hypothesis $\mathrm{\Delta }\ne 0$, if $f(x,y)=0$, then

therefore $2\mathit{ax}+\mathit{by}=0$ and $y=0$, so $x=y=0$, which implies that $f$ is definite, and this is a contradiction.

We have proved that if $f$ is a positive semidefinite (and not definite) form, then $\mathrm{\Delta }=0$. Similarly, if $f$ is a negative semidefinite (and not definite) form, then $\mathrm{\Delta }=0$.

Conversely, assume that $\mathrm{\Delta }=0$.

if $a=0$, then $b=0$ because $\mathrm{\Delta }=0$, and $f(x,y)=c{y}^2$ is semidefinite, but not definite since $f(1,0)=0$.

If $a\ne 0$, then

Therefore $f$ is positive semidefinite if $a>0$, and negative semi definite if $a<0$. Moreover $f(-b,2a)=0$, where $(-b,2a)\ne (0,0)$, so $f$ is not definite. □