Let A be matrix. If $bi$ ($b$ is real) is eigenvalue then $-bi$ is also eigenvalue. Why can it alone exist? $bi$ is eigenvalue and $-bi$ is not eigenvalue.
My attempt at proving this: [eigenvalues are roots of characteristic equation. Any equation, complex roots exist in pair. But how to prove this?]
$Ax = bi x$ and say $$Ax \neq -bix \\ \implies Ax \neq -bix = - Ax \\ 2Ax \neq0 $$
It should lead to contradiction somehow. I could not see it.
Also $Ax \neq -bix, bix =Ax \neq -bix \implies 2bix \neq 0$ it should be possible right say of x = [1,2] then $bix = bi(1+2) $\neq$ 0 fine right?
Please help me in understanding this silly doubt