Why can't we have only one complex eigenvalue?

Question

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

Answer 1

If $A$ is a real matrix,

$$Ax=bx$$ implies by conjugation

$$Ax^*=b^*x^*$$

and $b^*$ is an Eigenvalue.

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In the complex case, you can set the Eigenvalues that you want in a diagonal matrix (it can be simply $\begin{bmatrix}i\end{bmatrix}$ !).

Answer 2

For real matrix we always have complex conjugate eigenvalues. Indeed recall that for a real matrix the trace is real and the trace is equal to the sum of eigenvalues.

Otherwise it is not necessarly the case.

Answer 3

This is true if your matrix has real entries. As you said, eigenvalues are roots of the characteristic polynomial $f(x)$, which will have real coefficients. The conjugate root theorem then gives you what you need.