I'm taking a linear algebra course and we're learning about eigenvalues.
We were given as homework to prove the known discriminant based solution(s) of a quadratic equation:
Let $F$ be a field s.t. $char(F)\neq2$, and $a,b,c\in F$ with $a\neq 0$.
Prove the known root-discriminant relation for $ax^2+bx+c=0$:
- if $b^2-4ac$ isn't a square in $F$ then there are no solutions for the equation in $F$.
- if $b^2-4ac$ is a square then the roots are: ${ \lambda }_{ 1,2 }=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } $
Is there any real "linear algebra" based solution here?
(As opposed to doing it algebraically – starting from the known solution and working my way back to a quadratic equation)
Thanks!
Best Answer
I think I understand the question.
We want to formulate a quadratic equation as a linear algebra problem. Somehow we need to get a square. The simplest way to do that is to construct something of the form $v^\dagger A v$.
Imagine $v=\begin{pmatrix} 1 \\ x\end{pmatrix}$. Then the quadratic equation $ax^2+bx+c=0$ is exactly
$$\begin{pmatrix} 1 & x\end{pmatrix} \begin{pmatrix} c & b/2 \\ b/2 & a\end{pmatrix} \begin{pmatrix} 1 \\ x\end{pmatrix}=0$$
What condition on a matrix allows this to be possible? Why is this condition equivalent to the existence of a solution, if the quadratic part of the equation is not trivial?