Let $p$ and $q$ be real numbers such that $x^2+px+q\ \not= \ 0 $ for all real x. Prove that if n is an odd positive integer , then $X^2+pX+qI_n\ \not= \ 0_n$ for real matrices $ X$ of order n×n.
I am not able to show my attempt because I don't have any idea where to start . Thanks for the help.
Guide:
Suppose $X^2+pX+qI =O$
Since $n$ is odd, it must have a real eigenvalue, multiply it with the corresponding eigenvector to obtain a contradiction. I wll leave the details to you.
Edit:
Since $n$ is odd, then the charactheristic equation, $\det(x I-X)=0$, which is an odd degree polynomial must have a real root, $\lambda$.
That is we can find a real number $\lambda$ such that $\det(\lambda I-X)=0$,
Hence if you consider the linear system, $(\lambda I-X)v=0$, we must be able to find a non-zero vector $v$ that satisfies the equation. That is we must have such $v$ such that $Xv=\lambda v$.
Now, suppose that we have $X^2+pX+qI=O$, post multiply the equation by $v$ and try to simplify the system.