If $A$ is a $n\times n$ real matrix, then always exist reals $k,\lambda$ such that $\det(A^2+kA+\lambda I)=0.$

Question

Show that if $A$ is a $n\times n$ matrix with real elements, then always exists reals $k,\lambda$ such that $$ \det(A^2+kA+\lambda I)=0. $$

Answer 1

If A has a real eigenvalue r, choose $k=-2r$ and $\lambda=r^2$ If A has a complex eignevalue $a+bi$, choose $k=-2a$ and $\lambda=a^2+b^2$

Answer 2

$\det(A^2 + k A + \lambda I)=0$ if and only if $A$ has an eigenvalue $\mu$ with $\mu^2 + k \mu + \lambda = 0$. If $a$ and $b$ are the real and imaginary parts of the complex number $\mu$, then $(\mu - a)^2 + b^2 = 0$. Thus you can take $k = -2a$ and $\lambda = a^2 + b^2$.